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Goshia [24]
3 years ago
11

In the right triangle shown, AC = BC and AB = 8V2.

Mathematics
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

Each leg is 8 units long

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

Any right triangle must satisfy the Pythagorean Theorem

so

c^2=a^2+b^2

where

c is the hypotenuse (greater side)

a and b are the legs (perpendicular sides)

In this problem we have

a=AC\\b=BC\\AC=BC\\c=AB=8\sqrt{2}\ units

substitute

(8\sqrt{2})^2=a^2+b^2

(8\sqrt{2})^2=2a^2

128=2a^2

a^2=64\\a=8\ units

therefore

AC=BC=8\ units

Each leg is 8 units long

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Try to get x alone on one side

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Subtract 12x from both sides  

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(12a^(2) - 3)/(2)*(2a + 1)^(-2)*((6)/((2a + 1)))^(-1)
Elena L [17]

Answer:

\frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\

Step-by-step explanation:

\frac{(12a^2-3)}{2*(2a+1)^{-2}}*\frac{6}{(2a+1)^{-1} }\\ = \frac{(12a^2-3)}{2 * \frac{1}{(2a+1)^{2} }} * \frac{6}{\frac{1}{2a+1}  }\\ =\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}\\=\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}

= \frac{(12a^2-3)(4a^{2} + 1 + 2 (2a)(1))}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{1} * \frac{3}{2a+1}\\= \frac{3(12a^2-3)(4a^{2} + 1 + 4a)} {2a+1}\\\\= \frac{3(12a^2(4a^{2} + 1 + 4a) - 3 (4a^{2}+1+4a)} {2a+1}\\= \frac{3(48a^{4}+12a^{2}+48a^{3}  - 12a^{2} -12 - 4a)} {2a+1}\\= \frac{144a^{4}+36a^{2}+144a^{3}  - 36a^{2} - 36 - 12a)} {2a+1}

= \frac{144a^{4}+36a^{2}+144a^{3}  - 36a^{2} - 36 - 12a} {2a+1}\\= \frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\

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3 years ago
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