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adoni [48]
3 years ago
5

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

wer a list of questions.
Part A: For what one value of are the perimeters of the quadrilaterals the same? (Hint: The perimeter of a quadrilateral is the sum of its sides.)

Part B: For what one value of are the areas of the quadrilaterals the same? (Hint: The area of a quadrilateral is the product of its base and height.)

Mathematics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

For the perimeters, x must be equal to 2.

For the areas, it is either undefined, or something.

Step-by-step explanation:

You can first find the perimeters for both sides.

For the left shape, we add the two sides of 6 and x + 4 to get x + 10.

Then we multiply x + 10 by 2 because there are 4 sides, and we only got 2 sides.

The perimeter of the first shape is 2x + 20.

The second shape can be solved by doing the same thing by adding 2 and 3x + 4 to get 3x + 6.

3x + 6 times 2 is 6x + 12.

The second perimeter is 6x + 12.

If both sides are supposed to be equal, then we can write these two expressions we solved for like:

6x + 12 = 2x + 20.

Subtraction property of equality

6x + 12 - 12 = 2x + 20 - 12

Simplify

6x = 2x + 8

Again

6x - 2x = 2x - 2x + 8

Simplify

4x = 8

Division property of equality

4/4x = 8/4

Simplify

x = 2

So if x = 2, the perimeters will be the same.

You can confirm this by plugging it back into either equation.

For the areas, we just multiply the length and width for both shapes, so we get

6(x+4)  =  2(3x+4)

Since they are supposed to be equal.

We simplify and get

6x + 24 = 6x + 8

We know this is false and is not possible, since we can remove the 6x because it is on both sides.

We also know that 24 is not equal to 8 (who thought!)

:D

24 ≠ 8

So it is undefined or whatever you call it.

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I got 2 questions.
alexandr402 [8]

Answer:

1. In triangle ABC,

AC = a = 10, BC =b= 7 and ∠ C = 90°

By cosine law,

AB^2 = 7^2 + 10^2 - 2\times 7\times 10 cos 90^{\circ}

AB^2 = 49 + 100 - 0

AB^2 = 149

AB= \sqrt{149}

Now, by the law of sine,

\frac{sin B}{10} = \frac{sin90^{\circ}}{\sqrt{149} }

sin B = \frac{10}{\sqrt{149} }

\angle B = 55.0079798014\approx 55.008^{\circ}

2. In triangle ABC,

∠B = 30°,  AB=c=10 and ∠C = 90°

∠A = 180°-(30+90)°=60°

\frac{AC}{10}=sin30^{\circ}

⇒ AC = \frac{10}{2} = 5

By Pythagoras,

CB^2 = AB^2 - AC^2=10^2 - 5^2 = 100 - 25 = 75

⇒ CB = \sqrt{75} =8.66025403784\approx 8.66



6 0
3 years ago
Read 2 more answers
Given: <br> ABCD is a rectangle.<br><br><br><br> If AC = 12, then BE =<br><br> 3<br> 4<br> 6
AleksAgata [21]
6
AC is a diagonal and so is BD.
E is the center/mispoint
BE is half of BD which is 12 since it's equal to AC
12/2... 6
8 0
3 years ago
Read 2 more answers
Select all the values that are equivalent to -(7/8)
Deffense [45]

Answer:

last option

the correct answer is

- (- 7 / - 8)

7 0
3 years ago
What is 8,489÷9.This is 50 points question
s2008m [1.1K]
943.222222222222222222 (the 2 repeats) and it is rounded to 943.
5 0
4 years ago
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Please help need it asap
Studentka2010 [4]

Answer:

195

Step-by-step explanation:

87+38+40=165

360-165=195

Hope this helps!

If not, I am sorry.

4 0
2 years ago
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