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DaniilM [7]
3 years ago
10

Solve for x: 9x + 10 = - 4x + 36 helppo pleasee !!!

Mathematics
2 answers:
Misha Larkins [42]3 years ago
6 0

Answer: The answer is B or 2.

Step-by-step explanation:

dem82 [27]3 years ago
3 0

Answer:

x=2

Step-by-step explanation:

9x+10=-4x+36\\9x+4x=36-10\\13x=26\\x=26/13\\x=2

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Answer: -2x+25

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A gallup poll of 1236 adults showed that 12% of the respondents believe that it is bad luck to walk under a ladder. consider the
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7 0
3 years ago
Assume the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons.
Rama09 [41]

The rate of change is 16.8 million tons per year

<em><u>Solution:</u></em>

Given that, the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons

<em><u>The average rate of change is given by formula:</u></em>

\text{ Average rate of change } = \frac{\text{ change in value }}{\text{ number of years }}

Value in 1950 = 12 million tons

Value in 1955 = 96 million tons

Change in value = value in 1955 - value in 1950

Change in value = 96 million - 12 million = 84 million tons

Number of years = 1950 to 1955 = 5 years

<em><u>Substitute the given values in formula,</u></em>

<em><u></u></em>\text{ Average rate of change } = \frac{84 \text{ million tons}}{5} = 16.8 \text{ million tons}<em><u></u></em>

Thus rate of change is 16.8 million tons per year

5 0
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

Amount ( A ) = $ 14550

Time ( T ) = 10 years

Rate ( R ) = ?

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

The sum of principal and interest is called an amount.

From the definition,

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

plug the values

⇒\sf{14550 = 6000 + Interest}

Swap the sides of the equation

⇒\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 \: }

Interest = $ 8550

<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
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