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3 years ago

What percentage is equivalent to 14/25

Mathematics

2 answers:

lozanna [386]3 years ago

7 0

Answer:

0.56

Step-by-step explanation:

Hope this has helped. Have a good day.

zepelin [54]3 years ago

4 0

Answer:

use a calculator to divide 14 by 25 and the first two decimal places should be your answer

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Emily got a new job that guarantees her at 6% raise every year.If she started out making $25,000,How long will it be before she

Nadya [2.5K]

Hello from MrBillDoesMath!

Answer: about 12 years ( approx 11.9 years)

Discussion:

Salary at ...

end of year 1: 25000 + (.06) * 25000

end of year 2: (salary at end of year 1) +

.06(salary at end of year 1) =

(25000 + (.06)*25000) +

.06 (25000 + (.06)*25000)

The bold faced terms are the same. Factor out the boldfaced term to

get

(25000 + .06*25000) ( 1 + .06)

Factor 25000 out to get

25000 * (1 + .06) * (1+ .06)

The last two terms multiple out to give (1 + .06)^2

so....

end of nth year salary is

25000 * ( 1 + .06) ^n

To double her salary we need to find "n" such that

50000 = 25000 ( 1.06) ^ n

Dividing both sides by 25000 gives

2 = 1.06 ^n

Taking the logarithm of both sides gives:

log2 = n * log (1.06) or

n = log2 / log(1/06)

This is approximately 0.3010 / 0.0253 or about 11.9 years.

Thank you,

MrB

3 0

3 years ago

we would multiply the divisor and dividend of 1.44 / 1.2 by what number to make the divisor a whole number

kaheart [24]

To make the divisor and the dividend a whole number they bothneed to be multiplied by 100

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3 years ago

At 2 hours before sunset, the temperature is 0°F. At 2 hours after sunset, the temperature is −4°F. Identify the slope, y-interc

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Answer:-8

Step-by-step explanation:

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3 years ago

Read 2 more answers

Write the numeral one hundred fifty five and nine hundred two thousandths

7nadin3 [17]

Answer:

0.15590200

Step-by-step explanation:

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3 years ago

Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

4 years ago