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3 years ago

How do simplify 4(8+2f)+5 using distributive property?

Mathematics

2 answers:

polet [3.4K]3 years ago

8 0

8f+37 is the answer to check:
4 times 8 is 32 + 4 times 2f is 8f..........then 32+8f+5 to 8f + 37 is your answer

Fittoniya [83]3 years ago

6 0

This shows the step-by-step process required to do so!
4(8+2f)+5f
32+(4*2f)+5f
32+(8f)+5f
32+13f
Using the order of algebraic exponents to get our final answer:
13f+7

I hope this helped!

-Alex

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GaryK [48]

The initial value is where the line intersects the y-axis. So, this does it at y = 2. The graph has a slope (rate of change) of 1, because for every x-value, the y-value goes up by exactly 1. So, the answer is C.

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Kristin wants to wrap a ribbon around the perimeter of a rectangle. The length is five less than twice the width. The perimeter

romanna [79]

The length of the rectangle is 51 inches while the width of the rectangle is 28 inches.

Since Kristin wants to wrap a ribbon around the perimeter of a rectangle, the perimeter of a rectangle P = 2(L + W) where L = length of rectangle and W = width of rectangle.

Given that the perimeter of the rectangle, P = 56 inches and the length of the rectangle, L is five less than twice the width of the rectangle, W we have that

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Substituting L into P, we have

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2(W - 5) = 56

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W = 23 + 5

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L = 2W - 5

L = 2(28) - 5

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So, the length of the rectangle is 51 inches while the width of the rectangle is 28 inches.

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2 years ago

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Ulleksa [173]

Answer:

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Two collinear points on a line are given in the table below:

katrin [286]

Answer:

$(4,3)$ and $(7,2)$ do not lie on the line

Step-by-step explanation:

Given

$(0,0)\ and\ (2,1)$

Required

Determine which points that are not on the line

First, we need to determine the slope (m) of the line:

$m = \frac{y_2 - y_1}{x_2- x_1}$

Where

$(x_1,y_1) = (0,0)$

$(x_2,y_2) = (2,1)$

So;

$m = \frac{y_2 - y_1}{x_2- x_1}$

$m = \frac{1 - 0}{2-0}$

$m = \frac{1}{2}$

Next, we determine the line equation using:

$y - y_1 = m(x -x_1)$

Where

$m = \frac{1}{2}$

$(x_1,y_1) = (0,0)$

$y - y_1 = m(x -x_1)$ becomes

$y - 0 = \frac{1}{2}(x - 0)$

$y = \frac{1}{2}x$

To determine which point is on the line, we simply plug in the values of x to in the equation check.

For $(4,2)$

$x = 4$ and $y =2$

Substitute 4 for x and 2 for y in $y = \frac{1}{2}x$

$2 = \frac{1}{2} * 4$

$2 = \frac{4}{2}$

$2=2$

This point is on the graph

For $(4,3)$

$x = 4$ and $y = 3$

Substitute 4 for x and 3 for y in $y = \frac{1}{2}x$

$3 = \frac{1}{2} * 4$

$3 = \frac{4}{2}$

$3 \neq 2$

This point is not on the graph

For $(7,2)$

$x = 7$ and $y = 2$

Substitute 7 for x and 2 for y in $y = \frac{1}{2}x$

$2 = \frac{1}{2} * 7$

$2 = \frac{7}{2}$

$2 \neq 3.5$

This point is not on the graph

$(\frac{4}{8},\frac{2}{8})$

$x = \frac{4}{8}$ and $y = \frac{2}{8}$

Substitute $\frac{4}{8}$  for x and  $\frac{2}{8}$ for y in $y = \frac{1}{2}x$

$\frac{2}{8} = \frac{1}{2} * \frac{4}{8}$

$\frac{2}{8} = \frac{1 * 4}{8 * 2}$

$\frac{2}{8} = \frac{4}{16}$

$\frac{1}{4} = \frac{1}{4}$

This point is on the graph

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3 years ago