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Gala2k [10]
3 years ago
12

Sarah was thinking of a number. Sarah doubles it, then adds 4 to get an answer of 97.7. What was the original number?

Mathematics
1 answer:
rosijanka [135]3 years ago
7 0

Answer:

46.85

Step-by-step explanation:

let the number be n then doubling it is 2n and

2n + 4 = 97.7 ( subtract 4 from both sides )

2n = 93.7 ( divide both sides by 2 )

n = 46.85

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denis-greek [22]

Answer:

$3,000 invested at 6%

$7,000 invested at 10%

Step-by-step explanation:

Maria had total $10,000 to invest.

Let x be the amount that Maria invested initially at 6% interest rate.

0.06x

Then she invested the remaining amount at 10% interest rate.

0.10(10,000 - x)

She received a total of $880 in interest.

0.06x + 0.10(10,000 - x) = 880

0.06x + 1000 - 0.10x = 880

-0.04x = 880 - 1000

-0.04x = -120

0.04x = 120

x = 120/0.04

x = $3,000

This is the amount that Maria invested initially at 6% interest rate.

The remaining amount is

10,000 - 3,000

$7,000

This is the remaining amount that she invested at 10% interest rate.

3 0
3 years ago
PLEASE HELP ME!!!!!!!
abruzzese [7]

Answer:

8y-11 5/8

Step-by-step explanation:

product means multiplacation so multiply the terms used, which in this case is 8 and y. 11 5/8 less means subtraction. Therefore the answer. (also what grade is this question for?)

7 0
3 years ago
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The total current of a parallel circuit is 9+j2 amps and the current of one component of the circuit is 3−j5 amps. What is the c
stealth61 [152]

Apply law of total resistance in parallel circuit

\\ \tt\hookrightarrow \dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}

\\ \tt\hookrightarrow \dfrac{1}{9+2J}=\dfrac{1}{3-5j}+\dfrac{1}{R_2}

\\ \tt\hookrightarrow \dfrac{1}{R_2}=\dfrac{1}{9+2J}-\dfrac{1}{3-5J}

\\ \tt\hookrightarrow \dfrac{1}{R_2}=\dfrac{3-5J-9-2J}{(9+2J)(3-5J)}

\\ \tt\hookrightarrow R_2=\dfrac{(9+2J)(3-5J)}{-6-7J}

4 0
3 years ago
For what values of a does the equation a+x= 1+xa^2 have no solution?
Anton [14]

Answer:

We have the equation

a+x= 1+xa^2

we leave the terms with x on the left side of the equation and the independent terms on the right side.

x-xa^2=1-a\\(1-a^2)x=1-a\\

resolving for x we have that

x=\frac{1-a}{1-a^2}

Since we need that the equation doesn't have solution, then it is necessary that the denominator of x be 0 and this occur when 1-a^2=0\\1=a^2\\a=\pm1

Then, for a=\pm1 the equation hasn't solution

7 0
3 years ago
9 squared ? 11 ? 8 ? 4 ? 1 = 60 The ? Can be +,-,x or d division
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