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2 years ago

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3.5.2 Test (CSI) Question 6 of 12 Use the area of the rectangle to find the area of the triangle. 5 ft 10 ft O A. The area of th

e triangle is 25 square feet, because the area of the rectangle is 50 square feet O B. The area of the triangle is 25 square feet, because the area of the rectangle is 25 square feet C. The area of the triangle is 50 square feet, because the area of the rectangle is 50 square feet. D. The area of the triangle is 50 square feet, because the area of the rectangle is 100 square feet. SUBMIT PREVIOUS e 99+ Type here to search

1 answer:

Jlenok [28]2 years ago

3 0

Answer:

c

Step-by-step explanation:

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goldenfox [79]

Answer:

$1. \quad\dfrac{1}{k^{\frac{2}{3}}}\\\\2. \quad\sqrt[7]{x^5}\\\\3. \quad\dfrac{1}{\sqrt[5]{y^2}}$

Step-by-step explanation:

The applicable rule is ...

$x^{\frac{m}{n}}=\sqrt[n]{x^m}$

It works both ways, going from radicals to frational exponents and vice versa.

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2 years ago

1)The original price of concert is $100. They are on sale for 21% off. How much will the tickets cost after the discount? 2)Ms.B

mel-nik [20]

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2 years ago

432,2=26<br>275,3=66<br>342,5=41<br>123,4=20<br>then <br>543,2=?

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Answer:

I think 543,2=24.

hope it helps

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2 years ago

Calculate the slope, or rate of change, of this line.<br><br> 1) 3<br> 2) 4<br> 3) 2<br> 4) 0.5

ASHA 777 [7]

Answer:

The slope for the line is 2

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2 years ago

Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

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2 years ago