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RUDIKE [14]
3 years ago
5

A system of equations is shown below:

Mathematics
2 answers:
ioda3 years ago
8 0

Answer:

Solution (8 , 17).

Step-by-step explanation:

Given : y = 3x − 7  and y = 2x + 1 .

To find : What is the solution to the system of equations.

Solution : We have given

y = 3x − 7  --------(1)

y = 2x + 1  ---------(2).

Plug the value of y of equation 2 in equation 1.

2x + 1 = 3x - 7.

Subtracting both sides by 1

2x + 1 - 1 = 3x - 7 - 1.

2x = 3x - 8.

On subtracting both sides by 3x.

2x -3x = - 8.

-x = -8

On Dividing both sides by -1

x = 8.

Now plug the x = 8 in eq 1.

y = 3(8) -7.

y = 24 - 7.

y = 17.

Solution (8 , 17).

Therefore, Solution (8 , 17).

PolarNik [594]3 years ago
4 0
X = 8 and Y = 17 so the answer is A (8, 17)
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3 years ago
What simple interest rate would allow $6000 to grow to an amount of $14550 in 10 years?
Harman [31]

Answer:

\boxed{ \bold{ \huge{  \boxed{ \sf{ \: 14.25 \: \% \: }}}}}

Step-by-step explanation:

Given,

Principal ( P ) = $ 6000

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<u>Finding </u><u>the </u><u>Interest</u>

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\boxed{ \sf{Amount =  \: Principal + Interest}}

plug the values

⇒\sf{14550 = 6000 + Interest}

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⇒\sf{6000 + Interest = 14550}

Move 6000 to right hand side and change its sign

⇒\sf{Interest = 14550 - 6000}

Subtract 6000 from 14550

⇒\sf{Interest = \: 8550 \: }

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<u>Finding </u><u>the </u><u>rate </u>

{ \boxed{ \sf{Rate =  \frac{Interest \times 100}{Principal \times Time}}}}

plug the values

⇒\sf{ Rate = \frac{8550  \times 100}{6000 \times 10} }

Calculate

⇒\sf{Rate =  \frac{855000}{60000} }

⇒\sf{Rate = 14.25 \: \% \: }

Hope I helped!

Best regards!!

8 0
3 years ago
Help on these precalc problems 14-20
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Pls help I’ll brainlest
mylen [45]
Y=5/2x

Or y=2.5x
Hope this helps
4 0
3 years ago
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