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3 years ago

Find the length of side a.

Mathematics

2 answers:

Masteriza [31]3 years ago

8 0

<em>1</em><em>2</em>

=> here's your solution

=> hypotenuse = 13

=> base. = 5

=> we need to find out perpendicular

=> so, we know. h^2 = b^2 + p^2

=> now putting value in this formula

=> p^2 = 13^2 - 5^2

=> p^2 = 169 - 25

=> p^2 = 144

=> p = √144

= >> p = 12

coldgirl [10]3 years ago

6 0

The length of side A is 194

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My name is Ann [436]

Answer:

1 - First question = 6.00 | Second Question = 0.06

2 - Didnt know

3 - 1360, from 1.36kg

4 - 0.3175

5 - 20.5

6 - 18

7 - 965606.4

Step-by-step explanation:

sorry I couldn't give you number 2, but here are the rest of the answers

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3 years ago

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Vanyuwa [196]

All of them.
This is what they will turn out to be
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D will end up being 9x^2y^4

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3 years ago

Raymond spends 1/10 of his vacation budget on a car rental. Write an expression for how much Raymond has to spend on the rest of

Greeley [361]

Answer:

Step-by-step explanation:

1/10th of 650 is 65. then, do 650 - 65. 9/10th of 650 is 585, so the answer to this question is C.

6 0

3 years ago

Read 2 more answers

The figure shown below is composed of a semicircle and a non-overlapping equilateral triangle and contains a hole that is also c

Sloan [31]

Check the picture below.

since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle. We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.

now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.

$\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}$

$\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01$

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4 years ago

Enter the value, as a mixed number in simplest form, in the box. |2 1/3| = ?

Fittoniya [83]

Unclear of the question, but the value is already as a mixed fraction. The value of absolute value is just the distance from zero. So, |-4|=4 or |6|=6
$|2 \frac{1}{3} | = 2 \frac{1}{3}$

7 0

3 years ago