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3 years ago

Right triangle ACB has side lengths c = 17 inches and a = 8 inches

1 answer:

5 0

A²+b²=c²

Where a=leg, b=base, and c=hypotenuse.

So, because we are given the side lengths of c and a, we simply plug everything in, and try to find b.

a²+b²=c²

8² + b² = 17²

Simplify.

64 + b² = 289

Subtract 64 from both sides.

b² = 289 - 64

Simplify.

b² = 225

b = $\sqrt{225}$

So, b = 15 inches

~Hope I helped!~

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vichka [17]

Answer:

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2 years ago

Whoops i forgot to read the lesson :P 30 points

andriy [413]

Step-by-step explanation:

I'll do 2.

Alright,Alex let say we have factored a quadratic into two binomial, for example

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This means that

$if \: ab = 0 \: then \: a = 0 \: and \: b = 0$

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6 0

2 years ago

Don’t answer if you don’t know, plz. Thanks.

horrorfan [7]

B. I did this test too.

7 0

2 years ago

Maristella is constructing an isosceles triangle to use as a model in her Algebra class. The perimeter of her triangle is 24 cm.

Vlada [557]

Isolating s, the equation written in terms of s is:

$s = 12 - 0.5b$

-------------------------

The equation for the length of the third side, as stated in the problem,("Maristella uses the equation b = 24 – 2s to find b, the length of the triangle's third side") is:

$b = 24 - 2s$