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Rom4ik [11]
2 years ago
14

If the simple interest on ​$6000 for 9 years is ​$​2160, then what is the interest​ rate?

Mathematics
1 answer:
NISA [10]2 years ago
5 0

Step-by-step explanation:

<h2> l=P×R×T</h2><h3> 6000=2160×X/100×9</h3><h3> 6000÷194.4</h3><h2> =30.86%</h2>
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5 times 60 equals 5 times how many tens
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5*12 = 60 so 6 tens ig
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6:55 A. M. is the latest time he can leave his house

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6 0
3 years ago
Multiply 3/sqrt17- sqrt2 by which fraction will produce an equivalent fraction with rational denominator
zzz [600]

Answer:

B.

Step-by-step explanation:

To simplify something that looks like \frac{\text{whatever}}{\sqrt{a}-\sqrt{b}} you would multiply the top and bottom by the conjugate of the bottom. So you multiply the top and bottom for this problem I just made by:

\sqrt{a}+\sqrt{b}.

If you had  \frac{\text{whatever}}{\sqrt{a}+\sqrt{b}}, then you would multiply top and bottom the conjugate of \sqrt{a}+\sqrt{b} which is \sqrt{a}-\sqrt{b}.

The conjugate of a+b is a-b.

These have a term for it because when you multiply them something special happens.  The middle terms cancel so you only have to really multiply the first terms and the last terms.

Let's see:

(a+b)(a-b)

I'm going to use foil:

First:  a(a)=a^2

Outer: a(-b)=-ab

Inner:  b(a)=ab

Last:    b(-b)=-b^2

--------------------------Adding.

a^2-b^2

See -ab+ab canceled so all you had to do was the "first" and "last" of foil.

This would get rid of square roots if a and b had them because they are being squared.

Anyways the conjugate of \sqrt{17}-\sqrt{2} is

\sqrt{17}+\sqrt{2}.

This is the thing we are multiplying and top and bottom.

3 0
3 years ago
Read 2 more answers
If f(x)=x^2+3x+4 which of the number choices represents (f(x+h)-f(x))/h?
Helen [10]

In case you're not already aware, the expression \frac{f(x+h)-f(x)}h is called the "difference quotient" and represents the average rate of change of a function f over an interval [x,x+h].

For the function f(x)=x^2+3x+4, by substituting x+h we get

f(x+h)=(x+h)^2+3(x+h)+4=(x^2+2xh+h^2)+(3x+3h)+4=x^2+3x+4+(2x+3)h+h^2

Then the difference quotient is

\dfrac{f(x+h)-f(x)}h=\dfrac{(x^2+3x+4+(2x+3)h+h^2)-(x^2+3x+4)}h

=\dfrac{(2x+3)h+h^2}h=2x+3+h

where the last equality holds as long as h\neq0.

5 0
3 years ago
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