A bicycle wheel travels 321 in for each revolution. What is the radius of the wheel ​

Wind is caused by differences in the atmospheric pressure. When a difference in atmospheric pressure exists, air moves from the higher to the lower pressure area, resulting in winds of various speeds. On a rotating planet, air will also be deflected by the Coriolis effect, except exactly on the equator.

Hope this helped!

Answer:

24

Step-by-step explanation:

Round;

4.46 ≈ 4

5.75 ≈ 6

So;

4 x 6 = 24

Hope this helps!

Answer:

I think that the answer is 25

Step-by-step explanation:

I didn't really undertsand the question so I just multiplied 4.46 times 5.75 and than rounded to the nearest ones place.  The answer was 25.645

Hope this helped sorry if it is not the answer that you were looking for have a nice day.

The ground speed of the plane is 402 mph

We have given,

A plane leaves an airport traveling at 400 mph in the direction n 45° e. a wind is blowing at 40 mph in the direction n 45° w.

Now, Draw a right triangle with 400 and 40 as the two legs.

This is a right angle because 45 degrees NE is perpendicular to 45 degrees NW.

So, the two side lengths are 400 and 40

Using the Pythagorean theorem,

(a^2 + b^2 = c^2)

[tex]400^2+40^2=c^2\\\\c^2=160000+1600\\c^2=161600\\c=401.99[/tex]

Therefore the answer is 401.995, or 402mph.

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Answer:

101

Step-by-step explanation:

I time 41x2 than add 19

Answer: x is -2.5 or [tex]-2\frac{1}{2}[/tex]

Step-by-step explanation:

We need to solve for x

-3=4x+7

Step 1) Subtract 7 from both sides

-3-7=4x+7-7

-10=4x

Step 2) Divide both sides by 4 to isolate x

-10=4x

[tex]\frac{-10}{4} =\frac{4x}{4} \\-2.5=x[/tex]

Answer:

To find the interquartile range (IQR), ​first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

Step-by-step explanation:

[tex]Heyo![/tex]

SaddySokka is here to help!!

Let's do this step-by-step explanation!

[tex](15)(-5)+(15)(-3)[/tex]

[tex]=-75+(15)(-3)[/tex]

[tex]=-75+-45[/tex]

[tex]=-120[/tex]

Answer:

[tex]-120[/tex]

Hopefully, this helps you!!

Have a great day!!

SaddySokka~

Answer:

-120

Step-by-step explanation:

15×[-5]+15×[-3]

Use BODMAS

-75+-45

-120

Answer:

D

Step-by-step explanation:

(8x)² is the area of a square with a side length 8x,

because we want to find out how much space is inside a square that has two dimensions 8x length and 8x width.

Step-by-step explanation:

1)-1, 0, 1... do like this

The inequality model that can be used to solve the number of hours it would take Nada to drive is 2.5 ≤ x ≤3.

Average speed is the total distance travelled per time.

Average speed = total distance / total time

Time = total distance / average speed.

Time if she drives 150 miles = 150 / 60 = 2.5 hours

Time if she drives 180 miles = 180 / 60 = hours

2.5 ≤ x ≤3.

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Answer:

1^13 and 1^15

Step-by-step explanation:

1 raised to anything is still just 1

so, 1^13 = 1 and 1^15 =1

Answer:

Step-by-step explanation:

1.  x -> opposite side of 48°

   o → hypotenuse

   b → adjacent side of  48°

[tex]\sf Sin \ 48^\circ = \dfrac{opposite \ side }{hypotenuse}\\\\\\0.7431 = \dfrac{15}{o}\\\\\\0.74 * o = 15\\\\\\ o = \dfrac{15}{0.74}\\\\\\[/tex]

o = 20.27

[tex]\sf cos \ 48^\circ = \dfrac{adjacent \ side }{hypotenuse}\\\\\\0.67 =\dfrac{b}{o}\\\\\\0.67=\dfrac{b}{20.27}[/tex]

b = 0.67*20.27

b = 13.58

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2) i → opposite side of 25°

n → adjacent side of 25°

[tex]\sf Sin \ 25 =\dfrac{i}{t}\\\\\\0.42=\dfrac{i}{30}\\\\\\0.42*30=i[/tex]

i = 12.6

[tex]\sf Cos \ 30^\circ =\dfrac{n}{t}\\\\0.91=\dfrac{n}{30}\\\\\\0.91*30 = n[/tex]

n = 27.3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

3) a → opposite side of 70°

e → adjacent side of 70°

[tex]Sin \ 70^\circ =\dfrac{a}{l}\\\\0.94 =\dfrac{a}{25}\\\\0.94*25=a[/tex]

a = 23.5

[tex]\sf Cos \ 70^\circ =\dfrac{e}{l}\\\\0.34=\dfrac{e}{25}\\\\0.34*25=e[/tex]

e = 8.5

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

4)

[tex]\sf Sin \ 52^\circ = \dfrac{x}{75}\\\\0.79*75=x\\[/tex]

x = 59.25

[tex]\sf Cos \ 52^\circ = \dfrac{z}{75}\\\\0.62*75 =z[/tex]

z = 46.5

keeping in mind that vertical angles are always congruent.

[tex]\stackrel{\measuredangle A}{7x-6}~~ = ~~\stackrel{\measuredangle B}{8x-27}\implies -6=x-27\implies 21=x~\hfill \underset{\measuredangle B}{\stackrel{8(21)~~ - ~~27}{141}}[/tex]

Answer:

x ≤ -6

Step-by-step explanation:

The line is moving towards the left, which means that the numbers are getting smaller. X is going to be less than something.

The dot on -6 is colored in. This means that -6 is included in our inequality. [X could equal -6]

So, we could write the equation x ≤ -6

Answer:

W = 4 cm and  L = 9 cm

Step-by-step explanation:

I don't see a question, but will assume the problem wants the length(L) and width(W) of the described rectangle.

Let L and W stand for Length and Width.

Area of a rectangle is given by L*W

We are told that L*W = 36 cm^2

We are also told that L = 4W-7 ["length of a rectangle is 7 cm less than four times its width"]

Substituting the second into the first equation:

L*W = 36 cm^2

(4W-7)*W = 36 cm^2          [L = 4W-7]

4W^2-7W - 36 cm^2 = 0

(W-4)(4W+9) = 0

The roots are:  4 and -(9/4)

We'll use the positive value:  W = 4

Since L = 4W-7:

   L = 4(4)-7

   L = 16-7

  L = 9 cm

Answer:

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]

General Formulas and Concepts:
Calculus

Differentiation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]

Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]

Derivative Rule [Basic Power Rule]:

Integration

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Integration Method: U-Substitution + U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx[/tex]

Step 2: Integrate Pt. 1

Step 3: Integrate Pt. 2

Identify variables for u-substitution and u-solve.

1st Integral

3rd Integral

Step 4: Integrate Pt. 3

Let's focus on the 3rd integral first.

Step 5: Integrate Pt. 4

Focus on the other 2 integrals and solve using integration techniques listed above.

1st Integral:

[tex]\displaystyle\begin{aligned}\int\limits^5_3 {e^{3x}} \, dx & = \frac{1}{3} \int\limits^5_3 {3e^{3x}} \, dx \\& = \frac{1}{3} \int\limits^{15}_9 {e^{u}} \, du \\& = \frac{1}{3} e^u \bigg| \limits^{15}_9 \\& = \frac{1}{3} \bigg( e^{15} - e^9 \bigg)\end{aligned}[/tex]

2nd Integral:
[tex]\displaystyle\begin{aligned}7 \int\limits^5_3 {\cos x} \, dx & = 7 \sin x \bigg| \limits^5_3 \\& = 7 \bigg( \sin 5 - \sin 3 \bigg)\end{aligned}[/tex]

Step 6: Integrate Pt. 5

∴ we have evaluated the integral.

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Answer:

1086950.36760

Formula's used:

[tex]\rightarrow \sf \int sin(ax+b)=-\dfrac{1}{a} cos(ax+b)+c[/tex]

[tex]\rightarrow \sf \int cos(ax+b)=\dfrac{1}{a} sin(ax+b)+c[/tex]

[tex]\rightarrow \sf \int \dfrac{1}{ax+b} =\dfrac{1}{a} ln|ax+b|+c[/tex]

[tex]\rightarrow \sf \int e^{ax+b}=\dfrac{1}{a} e^{ax+b} + c[/tex]

[tex]\rightarrow \bold{ ln|a| - ln|b| = ln|\frac{a}{b} | }[/tex]

Explanation:

[tex]\dashrightarrow \sf \int \left(e^{3x}+7cos\left(x\right)-3tan^3\left(x\right)\right)[/tex]

                        apply sum rule: [tex]\bold{\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}[/tex]

[tex]\dashrightarrow \sf \int \:e^{3x}dx+\int \:7\cos \left(x\right)dx-\int \:3\tan ^3\left(x\right)dx[/tex]

                Integrate simple followings first, using formula's given above

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-\int 3tan^3x[/tex]

                        Breakdown the component

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int tan^2x(tanx)[/tex]

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int (sec^2x-1)(tanx)[/tex]

===========================================================

for integration of [tex]\bold{\int (sec^2x-1)(tanx)}[/tex]

[tex]\dashrightarrow \int \dfrac{-1+u^2}{u}[/tex]

[tex]\dashrightarrow \sf \int \:-\dfrac{1}{u}+udu[/tex]

[tex]\dashrightarrow \sf - \int \dfrac{1}{u}du+\int \:udu[/tex]

[tex]\dashrightarrow -\ln \left|u\right|+\dfrac{u^2}{2}[/tex]

[tex]\dashrightarrow \sf-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}[/tex]

================================================= insert back

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\left(-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}\right)[/tex]   outcome after integrating

Now apply the given limits

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{3(5)}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{3(3)}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                                                                   simplify

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{15}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{9}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                          and group the variables

[tex]\sf \hookrightarrow \dfrac{e^{15}-e^9}{3}-\dfrac{3}{2\cos ^2\left(5\right)}+\dfrac{3}{2\cos ^2\left(3\right)}+7\sin \left(5\right)-7\sin \left(3\right)+3\ln \left(\dfrac{1}{\cos \left(5\right)}\right)-3\ln \left(-\dfrac{1}{\cos \left(3\right)}\right)[/tex]

value:

[tex]\sf \hookrightarrow 1086950.36760[/tex]

Answer:

The answer is B. [tex]\frac{2x+1}{x^{2}-7} ,x\neq \sqrt{7}[/tex]

Step-by-step explanation:

[tex]f(x)=2x+1\\g(x)=x^{2} -7[/tex]

Focusing on the denominator makes it equal to = 0

[tex]x^{2} -7=0\\x^{2} =7\\x=+\sqrt{7},-\sqrt{7}[/tex]

If the denominator equals zero then the equation ends up undefined. A number over zero, zero can not go divided into a number.

Answer:

its 13 or B

Step-by-step explanation:

The inequality x -4 > 2 uses a greater than symbol

All numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2

The inequality is given as:

x -4 > 2

Add 4 to both sides of the inequality

x - 4 + 4 > 2 + 4

Evaluate the sum

x > 6

The above means that only numbers greater than 6 are in the solution of the inequality.

Since the options are not given, I will give a general solution that all numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2

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circumference of circle: 2πr

area of circle: πr^2

Using it's concept, it is found that there is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem:

Hence:

[tex]p = \frac{17}{23}[/tex]

There is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.

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Answer: B. 76

Step-by-step explanation: A right triangle is 180 degrees. It has an angle of 90 since it is a right triangle. 90 + 14 = 104. 180 - 104 = 76

Answer:

B. 76 degrees

Step-by-step explanation:

EVERY triangle's angles add up to 180 degrees. We already know that since it's a right triangle, one of the angles equals 90 degrees (that's a right angle) and they give us the second angle measurement, 14 degrees. If we add those two angle measures together and subtract them from 180, we should get the measure of the third angle as our answer.

14 + 90 = 104

180 - 104 = 76

Therefore, the third and final angle in this right triangle equals 76, so your answer is B. I hope this helps! Have a lovely day!! :)

The function f(x) = (3 * 0.04 * x) + x is a linear function, and the value of f(850) is 952

The function is given as:

f(x) = (3 * 0.04 * x) + x

Substitute 850 for x

f(850) = (3 * 0.04 * 850) + 850

Evaluate the product

f(850) = 952

Hence, the value of f(850) is 952

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Answer:

The number of degrees the temperature drops°

Step-by-step explanation:

hope this helps

and hope this is the answer you was looking for

pls mark brainliest

The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).

BODMAS stands for B - Bracket, O - order of Power, D - Division, M - Multiplication, A - Addition, and S - Subtraction.

To Arrange the expressions below in order from least to greatest. place the least at the top and the greatest at the bottom

72 / 8 - 2 x (3 + 1) equals 1

(72 / 8) - 2 x 3 + 1 equals 4

72 / (8 - 2) x 3 + 1 equals 37

72 / (8 - 2) x (3 + 1) equals 48

Thus, The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).

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Answer:

a) [tex]d-11 = t[/tex]

b) 7

Step-by-step explanation:

Let [tex]d[/tex] = daisies

Let [tex]t[/tex] = tulips

a) 11 fewer tulips than daisies: [tex]d-11 = t[/tex]

b) Substitute 18 into [tex]d[/tex] and solve for [tex]t[/tex].

[tex]18-11= t[/tex]

[tex]7 = t[/tex]