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3 years ago

An equilateral triangle has a perimeter of 15x + 3. What is the length of each side of the triangle?

Mathematics

1 answer:

maw [93]3 years ago

6 0

An equilateral triangle have 3 congruent sides.

Divide the total (15x +3) by 3

(15x + 3)/3 = 5x + 1

Each side's length is 5x + 1

hope this helps

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Find the slope of the line

Law Incorporation [45]

Answer:

-2

Step-by-step explanation:

$y = - 2x$

the line is going down meaning its a negative and to find the slope do rise over run

4 0

2 years ago

Please answer the following question. Also please answer if you actually know it.

Iteru [2.4K]

Given : f(x)= 3|x-2| -5

f(x) is translated 3 units down and 4 units to the left

If any function is translated down then we subtract the units at the end

If any function is translated left then we add the units with x inside the absolute sign

f(x)= 3|x-2| -5

f(x) is translated 3 units down

subtract 3 at the end, so f(x) becomes

f(x)= 3|x-2| -5 -3

f(x) is translated 4 units to the left

Add 4 with x inside the absolute sign, f(x) becomes

f(x)= 3|x-2 + 4| -5 -3

We simplify it and replace f(x) by g(x)

g(x) = 3|x + 2| - 8

a= 3, h = -2 , k = -8

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3 years ago

Recognize whether the fractions are Cleveland right yes Arnel use fraction tiles or number lines

wolverine [178]

1. No sorry I just know #1

7 0

2 years ago

Determine the distance between (-5,2) and (-2,5). show your work. round to the nearest tenth if necessary.

statuscvo [17]

Answer

-2

Step-by-step explanation:

learn how to do your work f^gg0+

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2 years ago

Polygon ABCD is dilated, rotated, and translated to form polygon A′B′C′D′. The endpoints of AB are at (0, -7) and (8, 8), and th

Anton [14]

Rotation and translation are rigid transformations, they don't change figure sizes. Dilation change figure sizes increasing or decreasing them by scale factor.

First, find AB and A'B' by the formula:

$AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}= \sqrt{(8-0)^2+(8-(-7))^2}=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17,\\ \\A'B'=\sqrt{(x_{B'}-x_{A'})^2+(y_{B'}-y_{A'})^2}= \sqrt{(2-6)^2+(1.5-(-6))^2}=\sqrt{4^2+7.5^2}=\sqrt{16+56.25}=\sqrt{72.25}=8.5.$

As you can see AB=2A'B'. This means that the segment AB was decreased twice to form segment A'B'. Then the scale factor is 1/2.

8 0

2 years ago