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3 years ago

14

Please help me solve number 1. Urgent.

1 answer:

Neko [114]3 years ago

5 0

Answer:

x^-(3/2) which is (1)

Step-by-step explanation:

squared root of x³ is basically x^(3/2)

And if the question is 1/x^(3/2), then you can simply answer x^-(3/2) which is (1)

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20000 into saving account 6% compound interest for 10 years

zimovet [89]

Answer:

210,000‬

Step-by-step explanation:

20,000 * 6% = 1,200‬

20,000 + 1,200 = 21,000

21,000 * 10 = 210,000‬

Therefore your answer is 210,000‬

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3 years ago

-2(x – 3) - 2<br> simplify the expression

Inessa [10]

Answer:

-2x+4

Step-by-step explanation:

-2(x-3)-2

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3 years ago

70, 000 grams= kilograms

Vesnalui [34]

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3 years ago

Read 2 more answers

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 $-1\le \sin \beta \le1$

4 0

3 years ago

Pls answer number 11 :d i will mark brianliest

QveST [7]

Answer:

D. (3; -4)

Step-by-step explanation:

.............

3 0

3 years ago