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2 years ago

8

Is 0.3434434443 a rational number

1 answer:

Kobotan [32]2 years ago

8 0

Answer:

Yes ,0.3434434443 is a rational number.

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kodGreya [7K]

Answer:

$24.35

Step-by-step explanation:

We will use the compound interest formula provided to solve this problem:

$A=P(1+\frac{r}{n} )^{nt}$

<em>P = initial balance</em>

<em>r = interest rate (decimal)</em>

<em>n = number of times compounded annually</em>

<em>t = time</em>

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First, change 1% into a decimal:

1% -> $\frac{1}{100}$ -> 0.01

Since the interest is compounded monthly, we will use 12 for n. Lets plug in the values now:

$A=800(1+\frac{0.01}{12})^{12(3)}$

$A=824.35$

Lastly, subtract <em>A </em>from the principal to get the interest earned:

$824.35 - 800 = 24.35$

8 0

2 years ago

Can someone help me please

zubka84 [21]

It’s too small zoom in some I will be glad to assist you

7 0

3 years ago

timothy creates digital art prints for his art show.in 1 day he created 8 diffrent art prints.which equation best represents y,t

Ghella [55]

Answer:

Step-by-step explanation:

3 0

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

4 0

2 years ago

What is the standard form of the following equation?

scZoUnD [109]

y = 2(x - 1)² + 4

y = 2(x² - 2x + 1) + 4

y = 2x² - 4x + 2 + 4

y = 2x² - 4x + 6

7 0

2 years ago