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3 years ago

15

The value of a card decreased by $5.100 over 3 years on average how much did it value decrease each year

1 answer:

vfiekz [6]3 years ago

3 0

Wait I’m confused about 5.100 is that money ?

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Which three lengths could be the lengths of the sides of a triangle? Thanks

Nataly_w [17]

s1+s2>s3 where s1 and s2 are the 2 smaller sides

Choice C is the only 1 that fits

7+13 >18

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3 years ago

What is the result of isolating x2 in the equation below?

Alex73 [517]

Answer:

Step-by-step explanation:

X^2=-y^2+10y+5

4 0

3 years ago

9 out of 25 students are home sick with the flu. What percentage of the students are home sick?

Arturiano [62]

Answer:

36%

Step-by-step explanation:

9/25

multiply each by 4 to get 36/100

as a percent it would be 36%

5 0

3 years ago

Read 2 more answers

Please help me with this question

jek_recluse [69]

A is the correct answer

E is 90 degrees

ABE is 52 so EBD is 64
$(180 - 52) \div 2 = 128 \div 2 = 64$

so D is 26:
$180 - (90 + 64) = 180 - 154 \\ = 126$

good luck

3 0

3 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 $-1\le \sin \beta \le1$

4 0

3 years ago