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yulyashka [42]
3 years ago
13

Evaluate. -2 (3) (4) ​

Mathematics
2 answers:
kykrilka [37]3 years ago
4 0

-23 :)

sorry i can't expain anything well

zheka24 [161]3 years ago
3 0
-2(3)(4)= -2 x 3 x 4

-2 x 3 x 4
\/
-6 x 4
\/
-24
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If interest is compounded every 3 months, how often is it compounded?
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Answer:

c. quarterly

Step-by-step explanation:

To start with 1 year is equal to twelve months

3months out of 12 months will be;

3/12= 1/4

Here it is compounded quarterly and n=4  where n is the number of compoundings a year.

Lets study the compound interest formula;

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

where;

A=The ending amount

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t=total number of years

The number of compoundings in any one year  can be an interest compounded  yearly where n=1, semi-annually with n=2, quarterly where n=4, monthly where n=12, weekly and n=52 and finaly daily with n=365.

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joja [24]

Answer:

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Step-by-step explanation:

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3 years ago
In the system of equations below, which variable would it be easiest to solve for? x + 4 y = 14. 3 x + 2 y = 12 The easiest to s
Murljashka [212]

Answer:

<h2>The easiest to solve for is x in the first equation</h2>

Step-by-step explanation:

Given the system of equation, x + 4 y = 14. and 3 x + 2 y = 12, to solve for x, we can use the elimination method of solving simultaneous equation. We need to get y first.

x + 4 y = 14............ 1 * 3

3 x + 2 y = 12 ............ 2  * 1

Lets eliminate x first. Multiply equation 1 by 3 and subtract from equation 2.

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x = 14-12

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It can be seen that the easiest way to get the value of x is by using the first equation and we are able to do the substitute easily <u>because the variable x has no coefficient in equation 1 compare to equation 2 </u>as such it will be easier to make the substitute for x in the first equation.

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4 years ago
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Answer:

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