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2 years ago

9

Seven tenths is_________pieces of the whole

2 answers:

riadik2000 [5.3K]2 years ago

8 0

10*10=100
7*10=70
70/100

ale4655 [162]2 years ago

4 0

70 pieces
7/10 = x/100
10*10=100
7*10 = 70
70/100

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What do i do for this problem

Margaret [11]

<span>The answer would be 20.

The question is asking you to find the value of the expression for when c = 100, so we have to put 100 where c is in the equation. Given that fact, we start with c/5, then put the 100 in for c.
c/5 = ??
100/5 = 20. </span>

4 0

2 years ago

Read 2 more answers

This Is challenging I hope you can figure it out

max2010maxim [7]

Answer:

980 hers your proof so 170 * 7 =180

Step-by-step explanation:

3 0

2 years ago

Matt is saving to buy a new motorcycle. If he deposits $70 at the end of each month in an account that pays an annual interest r

gladu [14]

P = $70, p = 2.5% = 0.025
q = 1 + 0.025 / 12 = 1.002
Future value of a periodic deposit:
A = P · q · ( q^30 - 1 ) / ( q - 1 )
A = 70 · 1.002 · ( 1.002^30 - 1 ) / ( 1.002 - 1 )
A = $2,166
Answer:
He will have $2.166 in 30 months.

6 0

3 years ago

Read 2 more answers

the circular ripple caused by dropping a stone in a pond is increasing in area at a constant rate of 20 square meters per second

mojhsa [17]

Answer:

2/π ≈ 0.637 m/s

Step-by-step explanation:

The rate of change of area with respect to time is ...

A = πr²

dA/dt = 2πr·dr/dt

Filling in given values in the above equations, we can find r and dr/dt.

25π = πr² ⇒ r = 5

20 = 2π·5·dr/dt

dr/dt = 20/(10π) = 2/π . . . . meters per second

The radius is increasing at the rate of 2/π ≈ 0.637 meters per second.

4 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