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3 years ago

Help Please! (8th Grade Math) :)

Mathematics

2 answers:

Yanka [14]3 years ago

6 0

Answer:

c = 75 in

Step-by-step explanation:

The hypotenuse of a right triangle is found by the Pythagorean theorem:

<em>a² + b² = c²</em> (a and b are the legs, c is the hypotenuse)

Plug in: <em>45² + 60² = c²</em>

Square + square root: √(2025 + 3600) = c

Add: √5625 = c

Square root: 75 in = c (no negatives because we are finding distance)

PilotLPTM [1.2K]3 years ago

4 0

75 inches. You use the pythagorean theorem

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Read 2 more answers

Rewrite the expression in terms of the given function 1/1-sinx - sinx/1+sinx

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Your question seems a bit incomplete, but for starters you can write

$\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}$

Expanding where necessary, recalling that $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ , you have

$\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}$

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

$\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x$

Now, since $1+\tan^2x=\sec^2x$ , the final form could also take

$\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1$

or

$\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x$

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