Austin spends a winter day recording the temperature once every three hours for science class. At 9 am, the temperature was -1.9°F. Between 9am and noon, the temperature rose 11.3°F. Between noon and 3pm, the temperature dropped 7.9°F. Between 3pm and 6pm, the temperature dropped 12.7°F. What was the temperature at 6pm?

Answers

Answer 1

To find the temperature at 6pm, we need to start with the temperature at 9am and then add or subtract the changes in temperature that occurred during the day.

We know that the temperature at 9am was -1.9°F. Between 9am and noon, the temperature rose 11.3°F, so at noon the temperature was:

-1.9 + 11.3 = 9.4°F

Between noon and 3pm, the temperature dropped 7.9°F, so at 3pm the temperature was:

9.4 - 7.9 = 1.5°F

Between 3pm and 6pm, the temperature dropped 12.7°F, so at 6pm the temperature was:

1.5 - 12.7 = -11.2°F

Therefore, the temperature at 6pm was -11.2°F.


Related Questions

Select the correct answer.

consider this equation.
cos(ф) = [tex]\frac{8}{9}[/tex]
if ф is an angle in quadrant iv, what is the value of tan(ф)?

Answers

The value of tan(ф) when cos(ф) = 8/9 in quadrant IV is - √17/8

To solve for the value of tan(ф), we need to use the trigonometric identity: tan(ф) = sin(ф)/cos(ф).

Since ф is in quadrant IV, we know that the cosine value is positive (due to cosine being positive in the adjacent side of quadrant IV) and the sine value is negative (due to sine being negative in the opposite side of quadrant IV).

We are given the value of the cosine, which is cos(ф) = 8/9. To find the sine, we can use the Pythagorean identity: sin²(ф) + cos²(ф) = 1.

Plugging in the given value of the cosine, we get: sin²(ф) + (8/9)² = 1. Solving for sin(ф), we get sin(ф) = - √(1 - (64/81)) = - √(17/81) = - √17/9.

Now that we have the values of sin(ф) and cos(ф), we can substitute them into the tan(ф) equation: tan(ф) = sin(ф)/cos(ф) = (- √17/9)/(8/9) = - √17/8.

Therefore, the value of tan(ф) when cos(ф) = 8/9 in quadrant IV is - √17/8.

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Out of 300 people sampled, 33 received flu vaccinations this year. Based on this, construct a 95% confidence interval for the true population proportion of people who received flu vaccinations this year. Give your answers as decimals, to three places < p <

Answers

A 95% confidence interval for the true population proportion of people who received flu vaccinations this year is  0.067 < p < 0.133.

To construct a 95% confidence interval for the true population proportion of people who received flu vaccinations, we can use the formula:

CI = p ± z√((p(1-p))/n)

where:

CI is the confidence interval

p is the sample proportion (33/300 = 0.11)

z is the z-score associated with a 95% confidence level, which is approximately 1.96

n is the sample size (300)

Substituting the values, we get:

CI = 0.11 ± 1.96√((0.11(1-0.11))/300)

CI = 0.11 ± 0.043

CI = (0.067, 0.133)

Therefore, the 95% confidence interval for the true population proportion of people who received flu vaccinations is 0.067 < p < 0.133.

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4. A person wants to buy a car from Toyota Company. If the price of car
including VAT is Birr 5,000,000 then,
a) What is the price of car before VAT?
b) What is the value of VAT?

Answers

Answer:

Step-by-step explanation:a) To find the price of the car before VAT, we need to first calculate the percentage of VAT included in the price:

VAT% = (VAT / Total Price) x 100

where VAT% is the percentage of VAT, VAT is the value of VAT, and Total Price is the price of the car including VAT.

From the given information, we have:

Total Price = Birr 5,000,000

VAT% = 15% (assuming a VAT rate of 15% in Ethiopia)

Therefore, we can solve for the value of the car before VAT as follows:

Total Price = Car Price + VAT

Birr 5,000,000 = Car Price + 0.15Car Price

Birr 5,000,000 = 1.15Car Price

Car Price = Birr 4,347,826.09

So the price of the car before VAT is Birr 4,347,826.09.

b) To find the value of VAT, we can use the same formula as above and solve for VAT:

Total Price = Car Price + VAT

Birr 5,000,000 = Birr 4,347,826.09 + VAT

VAT = Birr 652,173.91

Therefore, the value of VAT is Birr 652,173.91.

On the first swing, the length of the arc through which a pendulum swings is 50 inches. the length of each successive


swing is 80% of the preceding swing. determine whether this sequence is arithmetic or geometric. find the length of the


fourth swing

Answers

The length of the fourth swing is 25.6 inches. The sequence is arithmetic or geometric.

The length of the arc through which a pendulum swings is 50 inches. To determine whether the sequence is arithmetic or geometric, and to find the length of the fourth swing, we will analyze the given information.

The length of the first swing is 50 inches. Each successive swing is 80% of the preceding swing. This means that to find the length of the next swing, we multiply the length of the current swing by 80% (or 0.8).

Since we are multiplying by a constant factor (0.8) to find the next term in the sequence, this is a geometric sequence, not an arithmetic sequence.

Now, let's find the length of the fourth swing.

1st swing: 50 inches
2nd swing: 50 * 0.8 = 40 inches
3rd swing: 40 * 0.8 = 32 inches
4th swing: 32 * 0.8 = 25.6 inches

The length of the fourth swing is 25.6 inches.

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Please help

jkl will be rotated 90° clockwise around the origin to form j'k'l
what are the coordinates of point j' after the rotation? ​

Answers

The coordinates of point j' after the rotation are (-y, x), where x and y are the original coordinates of point j

The coordinates of point j' after rotating point jkl 90° clockwise around the origin can be found by applying the following transformation:

j' = (cos(90°) * x - sin(90°) * y, sin(90°) * x + cos(90°) * y)

Since we are rotating 90° clockwise around the origin, we have cos(90°) = 0 and sin(90°) = 1, so the transformation simplifies to:

j' = (0 * x - 1 * y, 1 * x + 0 * y)

j' = (-y, x)

Therefore, the coordinates of point j' after the rotation are (-y, x), where x and y are the original coordinates of point j.

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The difference between the record high and low temperatures in Chicago, Illinois is 109°F. The record low temperature was -5°F. Use an equation to find the record high temperature.

Answers

The value of the record high temperature is 114°F

Using an equation to find the record high temperature.

From the question, we have the following parameters that can be used in our computation:

The difference between the record high and low temperatures in Chicago, Illinois is 109°F. The record low temperature was -5°F.

This means that

High - Low = 109

Substitute the known values in the above equation, so, we have the following representation

High - 5 = 109

Add 5 to both sides

High = 114

Hence, the record high temperature is 114°F

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Find the measure of the question marked arc (view photo )

Answers

The arc angle indicated with ? is derived as 230° using the angle between intersecting tangents.

What is an angle between intersecting tangents

The angle between two tangent lines which intersect at a point is 180 degrees minus the measure of the arc between the two points of tangency.

angle G = 180° - arc angle HF

arc angle HF = 180° - 50°

arc angle HF = 130°

so the arc angle indicated with ? is;

? = 360° - 130°

? = 230°

Therefore, using the angle between the intersecting tangents, the arc angle indicated with ? is 230°.

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A consumer group is investigating two brands of popcorn, R and S. The population proportion of kernels that will pop for Brand R is 0. 90. The population proportion of kernels that will pop for Brand S is 0. 85. Two independent random samples were taken from the population. The following table shows the sample statistics. Number of Kernels in Samples Proportion from Sample that Popped Brand R 100 0. 92 Brand S 200 0. 89 The consumer group claims that for all samples of size 100 kernels from Brand R and 200 kernels from Brand S, the mean of all possible differences in sample proportions (Brand R minus Brand S) is 0. 3. Is the consumer group’s claim correct? Yes. The mean is 0. 92−0. 89=0. 3. Yes. The mean is 0. 92 minus 0. 89 equals 0. 3. A No. The mean is 0. 92+0. 892=0. 905. No. The mean is the fraction 0. 92 plus 0. 89 over 2 equals 0. 905. B No. The mean is 0. 92−0. 892=0. 15. No. The mean is the fraction 0. 92 minus 0. 89 over 2 equals 0. 15. C No. The mean is 0. 90+0. 852=0. 875. No. The mean is the fraction 0. 90 plus 0. 85 over 2 equals 0. 875. D No. The mean is 0. 90−0. 85=0. 5

Answers

The consumer group's claim that the mean of all possible differences in sample proportions (Brand R minus Brand S) is 0.3 is correct.

This can be calculated by subtracting the sample proportion of Brand S from the sample proportion of Brand R, resulting in a difference of 0.03 or 3%. This matches the consumer group's claim that the mean of all possible differences in sample proportions is 0.3. It is important to note that this result only applies to the specific samples taken and cannot be generalized to all samples from these brands.

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4) Your phone needs to be charged every other day and your tablet needs to be charged every third day. If you charge both today, how many days will it be until you need to charge both on the same day?​

Answers

To solve this problem, we need to find the least common multiple (LCM) of 2 and 3, which is the smallest number that is a multiple of both 2 and 3.

The multiples of 2 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, ...

The multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...

The smallest number that appears in both lists is 6, which is the LCM of 2 and 3. Therefore, you will need to charge both your phone and tablet on the same day again in 6 days.

Answer:

If they were both charged today, it would be on the sixth day that they were actually charging at the same time

A person was driving their car on an interstate highway


and a rock was kicked up and cracked their windshield


on the passenger side.


The driver wondered if the rock was equally likely to


strike any where on the windshield, what the probability


was that it would have cracked the windshield in his line


of site on the windshield. Determine this probability,


provided that the windshield is a rectangle with the


dimensions 28 inches by 54 inches and his line of site


through the windshield is a rectangle with the


dimensions 30 inches


by 24 inches.



a) 0. 373


b) 0. 139


c) 0. 423


d) 0. 476

Answers

There is about a 47.62% or 0.476 chance that the rock hit the windshield in the driver's line of sight. Option D.

To determine the probability that the rock hit the driver's line of sight on the windshield, we need to compare the area of the driver's line of sight rectangle to the total area of the windshield rectangle.

The area of the windshield rectangle is:

A1 = 28 in x 54 in = 1512 sq in

The area of the driver's line of sight rectangle is:

A2 = 30 in x 24 in = 720 sq in

Therefore, the probability that the rock hit the driver's line of sight on the windshield is:

[tex]P= \frac{A2}{A1}= \frac{720 \:sq in}{1512 \:sq in }[/tex] = 0.476 or 47.6%

So, there is about a 47.62% chance that the rock hit the windshield in the driver's line of sight.

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A lab technician is filling vitamin C capsules. He has 2.87 ounces of vitamin C and is putting 0.014 ounces of vitamin C into each capsule. How many capsules will the lab technician be able to fill with vitamin C? A. 3 B. 25 C. 402 D. 205 

Answers

Answer:

D) 205

Step-by-step explanation:

If the technician has a total of 2.87 oz, and can have a max of 0.014 oz in each capsule, we have to divide the total amount by the max amount per bottle.

2.87/0.014

=205

This means that the technician can fill 205 capsules with 0.014 oz of vitamin C.

Hope this helps!

Question 4 < > Evaluate ſtan® z sec"" zdz +C

Answers

To evaluate ſtan® z sec"" zdz +C, we can use integration by substitution. Let u = sec z, then du/dz = sec z tan z dz.

Using the identity 1 + tan^2 z = sec^2 z, we can rewrite the integral as:

∫ tan z (1 + tan^2 z) du

Simplifying this expression, we get:

∫ u^3 du

Integrating u^3 with respect to u, we get:

(u^4 / 4) + C

Substituting back u = sec z, we get:

(sec^4 z / 4) + C

Therefore, the solution to the integral ſtan® z sec"" zdz +C is (sec^4 z / 4) + C.
It seems like you are looking for the evaluation of an integral involving trigonometric functions. Your integral appears to be:

∫tan^n(z) * sec^m(z) dz + C

To solve this integral, we need the values of n and m. Please provide these values, and I'll be glad to assist you further in evaluating the integral.

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Evaluate the definite integral
∫ (t^5 - 2t^2)/t^4 dt

Answers

To evaluate the definite integral of the given function, ∫ (t^5 - 2t^2)/t^4 dt, follow these steps:
1. Simplify the integrand: Divide each term by t^4.
  (t^5/t^4) - (2t^2/t^4) = t - 2t^(-2)

2. Integrate each term with respect to t.
  ∫(t dt) - ∫(2t^(-2) dt) = (1/2)t^2 + 2∫(t^(-2) dt)

3. Apply the power rule to the remaining integral.
  (1/2)t^2 + 2(∫t^(-2+1) dt) = (1/2)t^2 + 2(∫t^(-1) dt)

4. Integrate t^(-1) with respect to t.
  (1/2)t^2 + 2(ln|t|)

Now, since we need to evaluate the definite integral, we should have the limits of integration. Let's assume the limits of integration are a and b. Then, apply the Fundamental Theorem of Calculus:

[(1/2)b^2 + 2(ln|b|)] - [(1/2)a^2 + 2(ln|a|)]

This expression gives the value of the definite integral for the given function within the limits a and b.

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Mollie drew mol and ted drew ted. they measured a few parts of their triangles
and found that ml = td, ol = ed, and l = d. what postulate can mollie and ted
use to justify why their triangles must be congruenta

Answers

Mollie and Ted can use the Side-Side-Side (SSS) postulate to justify why their triangles must be congruent.

According to the given information, the two triangles share three corresponding sides of equal length: ML = TD, OL = ED, and L = D.

The SSS postulate states that if three corresponding sides of two triangles are congruent, then the triangles are congruent. Therefore, because Mollie's triangle and Ted's triangle share three corresponding sides of equal length, they are congruent by the SSS postulate.

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Answer the following questions for the function f(x) = x Sqrt (x^2 + 4) defined on the interval - 5 ≤ x ≤ 5. f(x) is concave down on the interval x = to x =
f(x) is concave up on the interval x = to x = The inflection point for this function is at x = The minimum for this function occurs at x = The maximum for this function occurs at x =

Answers

The function f(x) = x Sqrt (x^2 + 4) is concave down on the entire interval. The inflection point is at x is equal to 0. There is no minimum or maximum in the interval.

To find the concavity, we need to find the second derivative

f(x) = x√(x^2+4)

f'(x) = √(x^2+4) + x^2/√(x^2+4)

f''(x) = -4x^3/(x^2+4)^(3/2)

The second derivative is negative for all x, which means the function is concave down on the entire interval.

To find the inflection point, we need to solve

f''(x) = 0

-4x^3/(x^2+4)^(3/2) = 0

This is true only when x = 0. Therefore, the inflection point is at x = 0.

To find the minimum and maximum, we need to find the critical points. The critical points are found by setting the first derivative equal to zero

f'(x) = √(x^2+4) + x^2/√(x^2+4) = 0

Multiplying both sides by √(x^2+4), we get

x^2+4 + x^2 = 0

2x^2 = -4

x^2 = -2

This equation has no real solutions, which means there are no critical points in the interval. Therefore, there is no minimum or maximum in the interval.

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Olympiads School Calculus Class 9 Test 1 -. Find the equation(s) of the tangent line(s) to the curve defined by x² + x²y2 + y = 1 when = -1. (4 marks) . Find the intervals of concavity and any point(s) of inflection for f(x) = x? In x. (4 marks)

Answers

The equation of the tangent line to the curve x² + x²y² + y = 1 at the point where x=-1 is (-dx/dy + 2)/(1 - √5)(x + 1). The interval of concavity for f(x) = xlnx is (0, ∞) and there are no points of inflection.

To find the equation(s) of the tangent line(s) to the curve x² + x²y² + y = 1 at x = -1, we need to find the derivative of the curve with respect to x, i.e.,

2x + 2xy²(dx/dy) + dy/dx = 0

At x = -1, we get

-2 + 2y²(dy/dx) + dx/dy = 0

dy/dx = (-dx/dy + 2)/(2y²)

Now, substituting x = -1 in the curve, we get

1 - y + y² = 0

Solving for y, we get

y = (1 ± √5)/2

Substituting y = (1 + √5)/2 in the equation for dy/dx, we get

dy/dx = (-dx/dy + 2)/(2(1 + √5)/4) = (-dx/dy + 2)/(√5 + 1)

Therefore, the equation of the tangent line to the curve at x = -1, y = (1 + √5)/2 is

y - (1 + √5)/2 = (-dx/dy + 2)/(√5 + 1)(x + 1)

Similarly, substituting y = (1 - √5)/2 in the equation for dy/dx, we get

dy/dx = (-dx/dy + 2)/(1 - √5)

Therefore, the equation of the tangent line to the curve at x = -1, y = (1 - √5)/2 is

y - (1 - √5)/2 = (-dx/dy + 2)/(1 - √5)(x + 1)

To find the intervals of concavity and any point(s) of inflection for f(x) = xlnx, we need to find the second derivative of the function with respect to x, i.e.,

f''(x) = (d²/dx²)(xlnx) = d/dx(lnx + 1) = 1/x

Now, to find the intervals of concavity, we need to find the values of x for which f''(x) > 0 and f''(x) < 0. We have

f''(x) > 0 when x > 0, which means the function is concave up on (0, ∞).

f''(x) < 0 when x < 0, which means the function is concave down on (0, ∞).

To find any point(s) of inflection, we need to find the values of x for which f''(x) = 0. However, in this case, f''(x) is never equal to zero. Therefore, there are no points of inflection for the function f(x) = xlnx.

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x+2y=6
-7x+3y=-8 (using substitution)

Answers

Answer:

point form - (2,2)

x=2 y=2

HELP!!!

Find the Area of a Rectengle with the base of 3x+1 in and a height of 2x-3 in.

A.5x^2-2 in^2

B.6x^2+7x-2 in^2

C.10x-4 in

D.6x^2-7x-3 in^2

Answers

The area of a rectangle with base of 3x+1 in and a height of 2x-3 in is given as follows:

D. A = 6x² - 7x - 3 in².

How to obtain the area of a rectangle?

The area of a rectangle of length l and width w is given by the multiplication of dimensions, as follows:

A = lw.

The dimensions for this problem are given as follows:

w = 3x + 1.l = 2x - 3.

Hence the expression for the area of the rectangle is given as follows:

A = (3x + 1)(2x - 3)

A = 6x² - 9x + 2x - 3

A = 6x² - 7x - 3 in².

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Which of the following is equivalent to [tex]\sqrt{x} 12qr^{2}[/tex]

Answers

The calculated value of the expression that is equivalent to √(x¹²qr²) is x⁶r√q

Calculating the expression that is equivalent to √(x¹²qr²)

From the question, we have the following parameters that can be used in our computation:

√x12qr²

Express properly

So, we have

√(x¹²qr²)

Evaluating the expression in the brackets using the law of indices

So, we have

√(x¹²qr²) = x⁶r√(q)

Next, we open the brackets

This gives

√(x¹²qr²) = x⁶r√q

Hence, the expression that is equivalent to √(x¹²qr²) is x⁶r√q

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8 Real / Modelling An advertising company uses a graph of this
equation to work out the cost of making an advert:
y=10+0.5x
where x is the number of words and y is the total cost of the bill in
pounds.
a)Where does the line intercept the y-axis?
b)How much is the bill when there are no words in the advert?
c)What is the gradient of the line?
d)How much does each word cost?

Answers

The gradient in the given equation is 0.5.

The given linear equation is y=10+0.5x where x is the number of words and y is the total cost of the bill in pounds.

a) When x=0, we get y=10

So, at (0, 10) the line intercept the y-axis.

b) $10 is the bill when there are no words in the advert.

c) Compare y=0.5x+10 with y=mx+c, we get m=0.5

So, the gradient of the line is 0.5

d) From equation, we can see the cost of each word is $0.5.

Therefore, the gradient in the given equation is 0.5.

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using mad when do you see a moderate amount of overlap in these two graphs when the mad is $20

Answers

Lower Bound: 20 - 29.67 ≈ -9.67

Upper Bound: 20 + 29.67 ≈ 49.67.

How to solve

The conversion of MAD to the standard deviation for normal distributions can be obtained using the given relationship:

Standard Deviation (σ) = MAD / 0.6745

For both distributions, the MAD holds a value of $20, thus arriving at σ ≈ 29.67.

There are two normal distributions now defined by their parameters as follows:

Mean (µ1) = $100 and Standard Deviation (σ1) = 29.67

Mean (µ2) = $120 and Standard Deviation (σ2) = 29.67.

Since both distributions share an equivalent standard deviation, we can perform a comparison of means to determine the overlap between them.

4

Typically there is observed moderate overlapping within one standard deviation from the difference in means.

The calculation of the difference in means indicates µ2 - µ1 = 120 - 100 = 20. Taking one standard deviation (which equates to 29.67) into consideration with respect to the difference of the means leads us to this range:

Lower Bound: 20 - 29.67 ≈ -9.67

Upper Bound: 20 + 29.67 ≈ 49.67.

It's noteworthy that negative values would not make sense within this context leading us to assume that the approximate overlap range is situated between $0 and $50 resulting in these normal distributions manifesting a sensible amount of overlap therein.

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The Complete Question

When comparing two normal distribution graphs with a mean of $100 and $120 respectively, and both having a MAD of $20, at what range do you see a moderate amount of overlap between the two distributions?

How do I find the answer to this problem The period of y = sin3 is _____

Answers

The period of the function y = sin(3x) is (2π/3).

I assume you meant to write "y = sin(3x)".

The period of the function y = sin(3x) can be found using the formula:

period = 2π / b

where "b" is the coefficient of x in the function.

In this case, b = 3, so:

period = 2π / 3

Therefore, the period of the function y = sin(3x) is (2π/3).

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-8
Find the distance, d, of AB.
A = (-7, -7) B = (-3,-1)
-6 -4
A
-2
B -2
-4
-6
-8
d = √x2-x1² + y2 - Y₁|²
d = [?]
Round to the nearest tenth.
Distance

Answers

Step-by-step explanation:

Using the distance formula:

d = √[(x2 - x1)² + (y2 - y1)²]

where A = (x1, y1) and B = (x2, y2), we can find the distance between points A and B as follows:

d = √[(-3 - (-7))² + (-1 - (-7))²]

d = √[4² + 6²]

d = √52

d ≈ 7.2

Therefore, the distance between points A and B is approximately 7.2 units, rounded to the nearest tenth.

Find the moment of inertia about the​ y-axis of the​
first-quadrant area bounded by the curve y=9−x^2
and the coordinate axes find ly (answer as a fraction)

Answers

To find the moment of inertia about the y-axis of the first-quadrant area bounded by the curve y=9−x^2 and the coordinate axes, we can use the formula:

I = ∫y² dA

where I is the moment of inertia, y is the distance from the y-axis to the infinitesimal element of area dA, and the integral is taken over the first-quadrant area.

To set up the integral, we need to express y in terms of x for the curve y=9−x². Solving for y, we get:

y = 9 - x²

The area element dA is given by:

dA = y dx

Substituting y in terms of x, we get:

dA = (9 - x²) dx

Now we can express the moment of inertia as an integral:

I = ∫y² dA
 = ∫(9 - x²)² dx     (limits of integration: x = 0 to x = 3)

To evaluate the integral, we can expand the integrand using the binomial theorem:

I = ∫(81 - 36x² + x⁴) dx
 = 81x - 12x³ + (1/5)x⁵     (limits of integration: x = 0 to x = 3)

Finally, we can substitute the limits of integration and simplify:

I = (81(3) - 12(3)³ + (1/5)(3)⁵) - 0
 = 243 - 108 + 27
 = 162

Therefore, the moment of inertia about the y-axis is 162 units^4.
To find the moment of inertia (Iy) about the y-axis for the first-quadrant area bounded by the curve y = 9 - x^2 and the coordinate axes, we need to integrate the expression for the moment of inertia using the limits of the region.

The curve intersects the x-axis when y = 0, so:

0 = 9 - x²
x² = 9
x = ±3

Since we're in the first quadrant, we're interested in x = 3.

The moment of inertia about the y-axis is given by the expression Iy = ∫x²dA, where dA is the area element. In this case, we'll use a vertical strip with thickness dx and height y = 9 - x². Therefore, dA = y dx.

Now, let's integrate Iy:

Iy = ∫x²(9 - x²) dx from 0 to 3

To solve this integral, you may need to use polynomial expansion and integration techniques:

Iy = ∫(9x² - x⁴) dx from 0 to 3
Iy = [3x³/3 - x⁵/5] from 0 to 3
Iy = (3(3)³/3 - (3)⁵/5) - (0)
Iy = (81 - 243/5)
Iy = (405 - 243)/5
Iy = 162/5

So the moment of inertia about the y-axis for the first-quadrant area bounded by the curve y = 9 - x^2 and the coordinate axes is Iy = 162/5.

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John earns $8. 50 per hour proofreading advertisements at a local newspaper. Write a function in function notation. Use d as your variable to represent days

Answers

The function notation is E(h) = 8.5h where h represents the number of hours worked so the domain is {0, 1, 2, 3, 4, 5} and the range is {0, 8.5, 17, 25.5, 34, 42.5}.

Let E(t) be John's earnings in dollars after working t hours, where t is in the domain 0 ≤ t ≤ 5. Then E(t) = 8.50t, since John earns $8.50 per hour proofreading ads.

The domain of the function is 0 ≤ t ≤ 5, since John works no more than 5 hours per day.

The range of the function is 0 ≤ E(t) ≤ 42.50 since John earns $8.50 per hour and works no more than 5 hours per day.

Therefore, the maximum earnings he can make in one day is 5 hours multiplied by $8.50 per hour, which equals $42.50.

The minimum earnings are $0, which would occur if John does not work at all.

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The question is -

John can earn $8.50 per hour proofreading adverse at a local newspaper. He works no more than 5 hours a day. Write a function in function notation and find a reasonable domain and range of his earnings.

Line x is parallel to line y. Line z intersect lines x and y. Determine whether each statement is Always True.

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Line x is perpendicular to line y. Line z crosses lines x and y. Only statements 3 and 4 are true.

∠6 = ∠8 is not true because they both lie on the same plane and makes an angle of 180° and can never be true. ∠6 = ∠1 is also not true because ∠1 is clearly obtuse angle and ∠6 is clearly acute angle so they cannot be equal. Hence, statement a and b are false.

∠7 = ∠3 is always true because they are corresponding angles and corresponding angles are always equal. m∠2 + m∠4 = 180° is also true because they lie on same plane and have common vertex and hence, they are supplementary angles and make a sum of 180°.  Hence, statement 3 and 4 is always true.

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There are four spaces each so you can put either parentasis or brakets

Answers

In the given function the domain is [-1, ∞]

Range is [-3, ∞]

The interval when function is positive [0,  ∞]

The domain of a function is the set of values that we are allowed to plug into our function.

This set is the x values in a function such as f(x).

The range of a function is the set of values that the function assumes

In the given function the domain is [-1, ∞]

Range is [-3, ∞]

The interval when function is positive [0,  ∞]

The interval when function is negative [-∞, -1]

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Determine whether the two figures are similar. If so, give the similarity ratio of the smaller figure to the larger figure. The figures are not drawn to scale.
*
Captionless Image
Yes; 3:5
Yes; 2:3
Yes; 2:5
No they are not similar

Answers

Determine whether the two figures are similar: D. No, the two figures are not similar.

What are the properties of quadrilaterals?

In Geometry, two (2) quadrilaterals are similar when the ratio of their corresponding sides are equal in magnitude and their corresponding angles are congruent.

Additionally, two (2) geometric figures such as quadrilaterals are considered to be congruent only when their corresponding side lengths are congruent (proportional) and the magnitude of their angles are congruent;

Ratio = 12/8 = 10/6 = 10/6

Ratio = 3/2 ≠ 5/3 = 5/3

In conclusion, the two figures are not similar because the ratio of their corresponding sides is not proportional.

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Missing information:

The question is incomplete and the complete question is shown in the attached picture.

Write a function to model the volume of a rectangular prism if the length is 26cm and the sum of the width and height is 32cm. what is the maximum possible volume of the prism?

Answers

To model the volume of a rectangular prism with length 26cm and width w and height h such that the sum of the width and height is 32cm, we can use the following function:

V(w, h) = 26wh

subject to the constraint:

w + h = 32

We can solve for one of the variables in the constraint equation and substitute it into the volume equation, giving us:

w + h = 32  =>  h = 32 - w

V(w) = 26w(32 - w) = 832w - 26w^2

To find the maximum possible volume, we can take the derivative of this function with respect to w and set it equal to zero

dv/dw= 832 - 52w = 0

Solving for w, we get:

w = 16

Substituting this value back into the constraint equation, we get:

h = 32 - w = 16

Therefore, the maximum possible volume of the prism is:

V(16, 16) = 26(16)(16) = 6656 cubic cm

So the function to model the volume of the rectangular prism is V(w) = 832w - 26w^2, and the maximum possible volume is 6656 cubic cm when the width and height are both 16cm.

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The area of a square with side length s is s2. Meg crocheted a baby blanket for her new cousin. The blanket is a square with 30-inch sides. What is the area of the baby blanket? Write your answer as a whole number or decimal

Answers

The area of the baby blanket with side length of 30 inches is equal to 900 square inches.

Let 'A' represents the area of the square.

And s represents the side length of the square.

The area of a square is given by the formula

A = s^2.

For Meg's baby blanket,

The side length of the baby blanket is equal to 30 inches,

Substitute the values in the area formula we get,

A = s^2

⇒ A = 30^2

⇒ A = 900 square inches

Therefore, the area of Meg's baby blanket is equal to 900 square inches.

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