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3 years ago

Hi! Can I please have help on solving linear graphing problems?

Mathematics

1 answer:

ElenaW [278]3 years ago

6 0

Answer:print out starter sheets

Step-by-step explanation:

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Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>

7 0

3 years ago

How do I solve 8x+3(x+5)-5(x-4)

adell [148]

Answer:

6x + 35

Step-by-step explanation:

First Distribute

8x + 3x + 15 -5x +20

Then collect like terms

6x + 35

to solve you would need the value of x to plug it!

7 0

3 years ago

Classify the triangle by its angles and sides

postnew [5]

Answer; A: Right scalene. Square on the corner tells you it's a right triangle. It's a scalene triangle because none of the sides are equal.

7 0

2 years ago

The sum of two consecutive integers is less than 55 the pair of integers with the greatest sum are 26 27

gayaneshka [121]

Answer:

Step-by-step explanation:

Let first integer = x

so second integer = x + 1

according to question we get the equation,,,,

(x) + (x + 1) < 55

x + x + 1 < 55

2x + 1 < 55

subtract 1 from both sides ,we get

2x < 54

divide both sides by 2 we get ......

x < 27

1st integer = x and x < 27

also as we need sum of integer with greates sum so we take x = 26

and 2nd integer = x+1

= 27 + 1

= 28

now if you add both the integers you get sum = 54 which is less than 55 and so this verify you second condtion also .......

hope this will help you

Please feel free to revert back for any further queries.

3 0

3 years ago

Write 10xy/15y2 in simplest form

jarptica [38.1K]

It would be 2x/3y assuming that’s meant to say 15y^2

3 0

3 years ago