Solve - 3g-2| +1 < 6.

Answer:

One litre of water has a mass of almost exactly one kilogram.kilograms = liters × density of a liquid.

Answer: rises; your slope is positive

Every single time your slope is positive, your line goes upward.

Every time your slope is negative, however, your line goes downward.

Rearrange 9x - 27y = 81 to -27y = -9x + 81 because we want to put our equation into slope intercept so that we can eventually get our 'y' all by itself.

Divide by -27 for every variable.

Your slope intercept form should now look like this; y = 1/3x - 3

Now that we're completely done or that we have 'y' by itself, it's safe to say that our answer is 1/3, which is, also, positive.

Hope this helps, dawg.

[tex]x^4+4x^3-8x-32=0[/tex]  

Can be factoried into:

[tex](x^3-8)(x+4)=0\\[/tex]

Now you're looking for which values of x will make the equations in the brackets equal to 0, since as long as one is equal to 0, they will be multiplied into 0.

First we'll do:

[tex]x^3-8=0[/tex]

We're looking for what number cubed is equal to 8, which is 2.

Note: -2 is also a possibility, but if you plug it in, you'll see that it doesn't work.

Second we'll do:

[tex]x+4=0[/tex]

This one's pretty easy, x will be -4.

So your solutions are:

x = 2, -4

The solutions of the polynomial [tex]x^{4}[/tex] + 4x³ - 8x - 32 = 0 are 2 and -4.

Polynomials are sums of terms of the form k⋅xⁿ.

where k is any number

n is a positive integer.

Example:

2x + 4x - 7 is a polynomial

We have,

[tex]x^{4}[/tex] + 4x³ - 8x - 32 = 0 _______(1)

Divide into two parts.

[tex]x^{4}[/tex] + 4x³

-8x - 32

Take out the common from each part.

x³ ( x + 4 )

- 8 ( x  + 4 )

Now,

We have,

x³ ( x + 4 ) - 8 ( x  + 4 ) = 0

We can write this as:

(x³ - 8) (x + 4) = 0

Now,

We need to find the x values.

x³ - 8 = 0

x³ = 8

x = ±2

Put x = 2 in (1)

[tex]x^{4}[/tex] + 4x³ - 8x - 32 = 0

x = 2

16 + 32 - 16 - 32

0 = 0

x = 2 is satisfied

Put x = -2 in (1)

16 - 32 + 16 - 32 = 0

-64 ≠ 0

x = -2 is not satisfied

x + 4 = 0

x = -4 putting in (1)

256 - 256 + 32 - 32 = 0

0 = 0

x = -4 is satisfied

Thus the solutions of the polynomial [tex]x^{4}[/tex] + 4x³ - 8x - 32 = 0 are 2 and -4.

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The patient consumed 935 milliliters of fluids in 24 hours.

Restrictions of fluid per day = 1000 milliliters

Consumption of fluid by patient in past 24 hours are :

    Milk = 3 ounces

IV fluid = 725 Milliliters

  Juice = 4 ounces

As we know that,

One fluid ounce = 30 milliliters

Then,     Milk = 3 × 30  = 90 milliliters

               Juice = 4 × 30 = 120 milliliters

To determine the total amount of fluids we will add the total amount of Milk, IV fluids and Juice.

Fluids consume by patient = 90 + 725 + 120

                                          = 935 milliliters

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Answer:

  A. 1/2 and 3/2

Step-by-step explanation:

You want to identify like fractions among the sets {1/2, 3/2}, {2/1, 2/3}, {10/10, 5/5}, and {7/4, 4/7}.

Like fractions are fractions that have the same denominator. Among the sets offered, the only one with denominators the same is ...

  {1/2, 3/2}

Here we go ~

A.) Derivative of 6x² :

[tex]\qquad \sf  \dashrightarrow \:f {}^{ \prime}(x) = \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{f(x + h) - f(x)}{h} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{6(x + h) {}^{2} - 6(x) {}^{2} }{h} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{6(x {}^{2} + 2xh + h {}^{2} ) {}^{} - 6(x) {}^{2} }{h} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ \cancel{6x {}^{2}} + 12xh +6 h {}^{2} {}^{} - \cancel{6x{}^{2}} }{h} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ 12xh +6 h {}^{2} {}^{} }{h} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ \cancel{ h}( 12x +6h ) {}^{} {}^{} }{ \cancel{h}} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: 12x + 6h[/tex]

[tex]\qquad \sf  \dashrightarrow \: 12x + 0[/tex]

[tex]\qquad \sf  \dashrightarrow \: f {}^{ \prime} (x) = 12x[/tex]

B.) The derivative of f(x) = 2x² -7x + 8 :

[tex]\qquad \sf  \dashrightarrow \:f {}^{ \prime}(x) = \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{f(x + h) - f(x)}{h} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ 2(x + h) {}^{2} - 7(x + h) + 8 - (2 {x}^{2} - 7x + 8)}{h} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ 2(x {}^{2} + 2xh + {h}^{2} ) {}^{} - \cancel{7x} + 7h+ \cancel8 - 2 {x}^{2} + \cancel{7x } - \cancel 8)}{h} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ \cancel{ 2x {}^{2}} + 4xh + 2{h}^{2} {}^{} - 7h - \cancel{2 {x}^{2}} }{h} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ \cancel {h}( 4x + 2{h}^{} {}^{} - 7) }{ \cancel{h} }[/tex]

[tex]\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: 4x + 2{h}^{} {}^{} - 7[/tex]

[tex]\qquad \sf  \dashrightarrow \: 4x - 0 - 7[/tex]

[tex]\qquad \sf  \dashrightarrow \: f {}^{ \prime} (x) = 4x - 7[/tex]

H.C.F of 3x³+15x² and 2x³-50x is x(x+5)

A number system is defined as a system of writing to express numbers.

The greatest common divisor of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers.

3x³ + 15x²

Factorize the given expression by taking out 3x²

3x²(x+5)

and the expression 2x³ - 50x ​

2x(x²-25)

2x(x+5)(x-5)

H.C.F of 3x³+15x² and 2x³-50x is x(x+5)

Hence, H.C.F of 3x³+15x² and 2x³-50x is x(x+5)

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Work Shown:

1/2 = 0.5

1/4 = 0.25

0.5 inch = 20 miles, which is the given scale

1 inch = 40 miles, after multiplying both sides by 2

3.25 inches = 130 miles after multiplying both sides by 3.25

The lengths of the other two sides of the triangle is 8√6 units.

Given that,

AD is the angle bisector of ∠A .

BD = 2 units .

DC = 4 units .

AE ⟂ BC and √15 units .

So, AB / AC = BD / DC { By angle bisector theorem }

AB / AC = 2/4

AB / AC = 1/2  ......... Eqn.(1)

Also, BC = BD + DC = 2 + 4 = 6 units .

Now let BE = x units .

So, in right-angled ∆AEB ,

AB = √(15 + x²) { By pythagoras theorem }

In right angled ∆AEC ,

AC = √{15 + (x - 6)²} { By pythagoras theorem }

Putting both values in Eqn.(1),

AB/AC = 1/2

√(15 + x²)/√(15 + x² + 36 - 12x) = 1/2

Squaring both sides,

(15 + x²) / (x² - 12x + 51) = 1/4.

4(15 + x²) = x² - 12x + 51.

60 + 4x² = x² - 12x + 51.

4x² - x² + 12x + 60 - 51 = 0.

3x² + 12x + 9 = 0.

3x² + 3x + 9x + 9 = 0.

3x(x + 1) + 9(x + 1) = 0.

(3x + 9)(x + 1) = 0.

x = (-3) and (-1) .

Taking x = (-1),

AB = √(15 + x²) = √(15 + 1) = 4 units .

So, AC = 4 * 2 = 8 units .

Taking x = (-3),

AB = √(15 + 9) = √24 = 4√6 units.

So, AC = 2 * 4√6

= 8√6 units .

Hence, the lengths of the other two sides of the triangle is  8√6 units.

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The inverse of the statement  p → q is "If x – 5 ≠  10 then

4x + 1 ≠  61"

The statement is given as

p: x – 5 =10

q: 4x + 1 = 61

The inverse of a statement is such that:

The hypothesis and conclusion are negated.

This means that a conditional statement referred to as p → q would have the inverse of -p → -q.

When the above rule is applied on the statement

p: x – 5 ≠  10

q: 4x + 1 ≠  61

This means that the inverse of the statement  p → q is "If x – 5 ≠  10 then

4x + 1 ≠  61"

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Answer:

A

MF up there wont tell u but its A

Based on the probability of rain in Nevada, the probability that it will not rain is 0.7.

The probability that if it does not rain, the school bus will be on time is 0.85.

The probability that it will rain and the school bus will still be on time is 0.6.

The probability that it will not rain in Nevada can be found by the formula:

= 1 - probability of rain in Nevada

= 1 - 0.3

= 0.7

The probability that if it does not rain, the school bus will be on time is:

= 1 - probability of the school bus being late if it does not rain

= 1 - 0.15

= 0.85

The probability that it will rain and the school bus will still be on time is:

= 1 - probability of the school bus being late if it rains

= 1 - 0.4

= 0.6

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Y equals f(x) is written on the first line, The second line, marked y = g(x). When the input is for either of the two functions, the result is x=0 and return the output is same value.

Given that,

Y equals f(x) is written on the first line, which also passes through (-1, 2), (0, 2), and (2, 2). The second line, marked y = g(x), points at (0, -1), passes through (-0.5, -2), and (1, 1). The lines come together at (1.5, 2).

We have to find the input value for which f(x) = g(x) is true using the graph. x = 0.5, x = 1, x = 1.5, x = 2.

The final two actions:

f(x)=-2/3(x+1)

g(x)=1/3(x-2)

The output are equation

f(x)=g(x)

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

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

-2x-2=x-2

-2x-x=-2+2

-x=0

x=0

When x = 0 is used as an input, the first two functions always return the same results.

f(x) traverses (-1,2), (0,2), and (2,2)

The paths that g(x) takes are (-0.5,-2), (0,-1), (1,1)

Therefore, When the input is for either of the two functions, the result is x=0 and return the output is same value.

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make it simple its x=1.5

Using z-scores, it is found that due to the lower z-score, Beth had the fastest time when compared to her team.

This problem is incomplete, but researching on the internet, we have that:

The z-score of a measure X in a distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

For this problem, lower times means that the swimmer is faster, hence the swimmer that had the fastest time when compared to her team is the swimmer with the lowest z-score.

Angie's z-score is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (26.2 - 27.2)/0.8

Z = -1.25.

Beth's z-score is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (27.3 - 30.1)/1.4

Z = -2.

Due to the lower z-score, Beth had the fastest time when compared to her team.

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The sinusoidal function for the given conditions can be written as h(t) = 9cos(1.67π(x - 2.1)) + 39.

The term sinusoidal refers to a curve, also known as a sine wave or a sinusoidal, that shows smooth, periodic oscillation. It is named after the function y=sin (x). Sinusoidal appear often in mathematics, physics, engineering, signal processing, and many other fields.

A general form of a cosine function is given as shown below,

g(x) = a cos(bx+c) + d

Where the values of the given constant is,

a = amplitude

b = The period is of 2pi/B

c = phase shift

d = vertical shift

Since for the given condition the greatest value is 48 and the smallest value is 30 (a difference of 18), therefore, the amplitude is of the function can be written as,

2a = 18

a = 9

Further, a conventional cosine function with amplitude 9 would fluctuate between -9 and 9, while this one ranges between 30 and 48, resulting in d = 39 vertical shift.

Also, The minimum and maximum values form half the period, therefore, we can write,

π/B = 2.7 - 2.1

B = π/0.6

B = 1.67π.

Furthermore, The maximum value in the normal function is at x = 0, but the greatest value in this function is at x = 2.1, thus, the phase shift is 2.1 units to the right, or c = -2.1.

Hence, the sinusoidal function for the given conditions can be written as h(t) = 9cos(1.67π(x - 2.1)) + 39.

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Answer: range = 16 standard deviation = 5

Step-by-step explanation:

Answer:

22,400

Step-by-step explanation:

80,000 x .28

The length of the major arc QSR is 39.77 inches

The given parameters are:

m∠QPR = 170 degree

Radius, r = 6 inches

The length of the major arc QSR is calculated as:

Arc length = (360 - Angle/360) * 2πr

Substitute the known values in the above equation

Arc length = (360 - 170/360) * 2 * 3.14 * 6

Evaluate the difference

Arc length = (190/180) * 2 * 3.14 * 6

Evaluate the product

Arc length = 39.77

Hence, the length of the major arc QSR is 39.77 inches

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The price per item is 765.

The variable cost of 100 units is 1500.

The fixed expense is 750.

The expense for 100 units is 2250

Fixed costs are costs that do not vary with output. e,g, rent, mortgage payments. Variable cost change with the unit of output.

The variable cost of 100 units = variable cost per unit x number of units

15 x 100 = 1500

The expense for 100 units = fixed cost + total variable cost

750 + 1500 = 2250

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Answer:

x=-4

Step-by-step explanation:

this is if by roots, the problem is asking for the zeros of the equation

The value of "x" which will satisfy the equation [{21090(x -1 ) + 109.07 } = 1] is 1.01.

As per the question statement, we are provided with an equation [tex][{21090(x -1 ) + 109.07 } = 1][/tex].

We are required to solve the above mentioned equation for "x", such that the value of x when substituted in the equation, satisfies the same.

To solve this question, we will simply use the methods of rearranging and opening up of brackets, as shown below.

[tex][{21090(x -1 ) + 109.07 } = 1] \\or, 21090x - 21090+109.07=1\\or, 21090x - (21090-109.07)=1\\or, 21090x - 20980.93=1\\or, 21090x=(1+20980.93)\\or, 21090x=20981.93\\or,x=\frac{21090}{20981.93}\\or,x=1.00515\\or,x=1.01[/tex]

Therefore, the value of "x" which will satisfy the equation

[{21090(x -1 ) + 109.07 } = 1] is 1.01.

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Aaron Smith would has average hits per baseball game as 0.79  hits per game.

Average provides ill information in case of skewed data.

Arithmetic mean is the best central measure available for representing the values of a data set.  It is also called average of the values of the considered data set. It serves as predicted value(in case no other information of the data is available) of that data set.

WE have been given that After 30 baseball games Aaron Smith had 24 hits. If after 100 games he had 80 hits, then he would has average hits per baseball game;

Average  = Hits in game / number of games

Average  = 24- 80 / 30-100

Average  =  56 / 70

Average  = 0.79  hits per game

Hence, Aaron Smith would has average hits per baseball game as 0.79  hits per game.

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A decimal point is used when working with dollars but the decimal point is not necessary when working with cent for each dollar amount, the equivalent amount expressed as cents of $5.74 is 574 cents and $0.16 is 16 cents

We know that 1 dollar is equal to 100 cents

So, x dollar = x(100) cents

For $5.74 = 5.74(100) cents

                 = 574 cents

For $0.16 = 0.16(100) cents

                 = 16 cents

Integer and non-integer numbers are represented using the decimal numeral system, also known as the base-ten positional numeral system, denary, or decanary. The Hindu-Arabic numeral system has been expanded to include non-integer values. Decimal notation is the term used to describe how numbers are represented using the decimal system.

It is indicated by a '.', for instance, 3.14

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The location of a center of dilation in the figure is at the point where the corresponding vertex of the pre–image and image overlaps, which is point N, the correct option is therefore;

The given pre–image = Parallelogram LMNO

The image obtained from the pre–image = Parallelogram L'M'N'O'

Required: The location of the center of dilation

Solution:

The center of dilation is the point about which the figure or image is dilated.

It is the point that does not change following the dilation.

A vertex on the pre–image that gives an image vertex at the same point, is at the center of dilation which does not change in both the pre–image and image.

Therefore, given that point N and N' coincides, which indicates that the distance the pre–image point N is dilated to get the image point, N is 0. The center of dilation is at the vertex N.

The correct option is therefore;

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Answer:

On a vertex of figure LMNO

Step-by-step explanation:

Answer: 46

Step-by-step explanation:

Since we know that angles J and E are congruent, we can do the equation 6x + 4 = 2x +32. We then solve for x which gives us 7. After that, we plug in the 7 into the equation for E which would be = 2(7) + 32. Once we solve that, we end up with 46.

Answer: 21x^3 + 3 + 12x^2 + x

Step-by-step explanation:

The average rate of change of f(x)=[tex]8x^{2} -5[/tex] on the interval [4,b] is [tex]\frac{128-8b^{2} }{4-b}[/tex]

Given,

The function f(x) =[tex]8x^{2} -5[/tex]

The intervals = [4,b]

The average rate of change = [tex]\frac{f(a)-f(b)}{a-b}[/tex]

Where a and b are the interval

f(4)= [tex]8(4)^{2}-5[/tex]

=123

f(b)= [tex]8b^{2}-5[/tex]

The average rate of change = [tex]\frac{123-(8b^{2}-5) }{4-b}[/tex]

[tex]=\frac{123-8b^{2}+5 }{4-b} \\=\frac{128-8b^{2} }{4-b}[/tex]

Hence, The average rate of change of f(x)=[tex]8x^{2} -5[/tex] on the interval [4,b] is [tex]\frac{128-8b^{2} }{4-b}[/tex]

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