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kodGreya [7K]
3 years ago
13

The length of a rectangle is

Mathematics
2 answers:
Aleks04 [339]3 years ago
5 0
5,184. The wording of the question is hard to understand (if im wrong)
notka56 [123]3 years ago
4 0
Your side lengths are 11, 11, 7, 7. Area of a rectangle is bh so the answer is 77in^2.
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find the missing angle measure in each triangle. Then classify the triangle as acute, right, or obtuse. PLEASE ANSWER QUICK ILL
larisa86 [58]
X=72 cause 180-(51+57) so 180-108 and the triangle is acute cause all of the angle measures are below 90 degrees.

So your answer is
x=72
Acute Triangle
8 0
3 years ago
Read 2 more answers
Given a triangle with: a =<br> 150, A = 75°, and C = 30°<br> Using the law of sines gives: c = 0
koban [17]

Answer:

c = 77.6

Step-by-step explanation:

You may have entered the measure of a side as the measure of an angle.

\dfrac{\sin A}{a} = \dfrac{\sin C}{c}

\dfrac{\sin 75^\circ}{150} = \dfrac{\sin 30^\circ}{c}

c\sin 75^\circ = 150 \sin 30^\circ

c = \dfrac{150 \sin 30^\circ}{\sin 75^\circ}

c = 77.6

You are correct. Good job!

5 0
3 years ago
Find the area of the shaded region
Alexandra [31]
Take the area of the whole shape and then subtract the area of the white part, and the shaded part's area will then be left.
(8*24)-(0.5*24*8)=96
Answer D
3 0
3 years ago
Read 2 more answers
There is 3/4 of a box of cereal remaining. You eat 2/5 of the remaining cereal. What fraction of the box do you eat?
Bumek [7]
5/8 of the box is right
6 0
4 years ago
7 * 0.04 = 7 * 4 *<br> 7 times 0.04 equal 7 times 4 times ?
wolverine [178]

Answer:

Solution

1 Cancel 77 on both sides

x\times 0.04=x\times 4xx×0.04=x×4x

2 Regroup terms

0.04x=x\times 4x0.04x=x×4x

3 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

​a

​​ x

​b

​​ =x

​a+b

​​

0.04x={x}^{2}\times 40.04x=x

​2

​​ ×4

4 Regroup terms

0.04x=4{x}^{2}0.04x=4x

​2

​​

5 Move all terms to one side

0.04x-4{x}^{2}=00.04x−4x

​2

​​ =0

6) Factor out the common term x

x(0.04-4x)=0

7 Solve for x

x=0,0.01

Done!

Step-by-step explanation:

6 0
3 years ago
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