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3 years ago

∠A is an acute angle in a right triangle.

1 answer:

3 0

Cosine = adjacent / hypotenuse
So that means the adjacent side = 15 and the hypotenuse = 17
By Pythagorean Theorem, we know that
opposite^2 = hypotenuse^2 - adjacent^2
opposite^2 = 17^2 - 15^2
opposite^2 = 289 - 225
opposite^2 = 64
opposite side = 8

sine = opposite / hypotenuse
So,
sine A = 8 / 17

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Find the volume of this prism, please

Elodia [21]

Answer
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Step by Step
8(length)x7(width)x14(height)=784 cubic inches

5 0

3 years ago

Read 2 more answers

Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)

Leni [432]

Alrighty

remember some rules
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}=\frac{1}{x^m}$
and
$x^0=1$ for all real values of x
and
$(a^b)^c=a^{bc}$
and
$(\frac{a}{b})^c=\frac{a^c}{b^c}$
and
(a/b)/(c/d)=(ad)/(bc)
and
$(a^b)(a^c)=a^{b+c}$
and
$\frac{a^m}{a^n}=a^{m-n}$
and don't forget pemdas
example: $-x^m=-1(x^m)$ but $(-x)^m=(-1)^m(x^m)$

so

4.
$(\frac{c^{-2}}{2})^{-2}=$

$\frac{(c^{-2})^{-2}}{2^{-2}}=$

$\frac{c^4}{\frac{1}{2^2}}=$

$\frac{c^4}{\frac{1}{4}}=$

$4c^4$

5.

$\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=$

$\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=$

$\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=$

$\frac{(a^4bc^5)b^3}{(-1)(a^2)}=$

$\frac{-a^4b^4c^5}{a^2}=$

$-a^2b^4c^5$

3 0

3 years ago

In another problem in my homework, it says look at problem 4, problem 4 has to do with the question about seans team when it had

notka56 [123]

I dont know because I’m confused

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3 years ago

The annual water usage in rockport increases at a rate of 10% every year as the town gets bigger. this year, the town used 6,740

Finger [1]

Answer:

Step-by-step explanation:

<h3> I=PRT/100</h3><h2> 6740×<u>1</u><u>0</u>×7</h2><h2> 100</h2><h2> 674×1×7</h2><h2> 4618</h2><h2>the nearest hundredth is a 46•18 this is my answer</h2>

6 0

2 years ago

Square root for 9/100

Leno4ka [110]

$\sqrt{a}=b\iff b^2=a\ for\ a\geq0\ and\ b\geq0.\\\\\sqrt{\dfrac{9}{100}}=\dfrac{\sqrt9}{\sqrt{100}}=\dfrac{3}{10}$

5 0

3 years ago