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spin [16.1K]
3 years ago
15

Solve the following equation for x.

Mathematics
2 answers:
Crazy boy [7]3 years ago
5 0
X=(-4) or X= 4
add 112 to both sides
divide both sides by 7
then take the square root of 16
eduard3 years ago
5 0

Answer:

<u>C. X= -4, X= 4.</u>

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What is the justification for the step taken from line 2 to line 3?
stiv31 [10]

Step-by-step explanation:

3x+9-7x=x+10+x\\-4x+9=2x+10\\-6x+9=10\\

  • In the line 2, only variables were operated with a reduction of similar terms in each side of the equation, giving as result -4x at one side, and 2x at the other side.
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3 0
3 years ago
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6% of a value is 510
Dennis_Churaev [7]

Answer:

The value is 8500

Step-by-step explanation:

Of means multiply and is means equals

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3 years ago
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4 0
3 years ago
A population of insects grows exponentially, as shown in the table. Suppose the increase in population continues at the same rat
Ivanshal [37]

We are told that a population of insects grows exponentially and we are given a table of data about insect population growth. We are asked to find population of insects at the end of week 11.        

The initial insect population is 20 and at the end of 1st week population increases to 30.

Let us find growth percentage of insect  population,

\text{Growth percentage}=\frac{\text{Difference}}{Actual} \cdot100

\text{Growth percentage}=\frac{\text{30-20}}{20} \cdot100

\text{Growth percentage}=\frac{\text{10}}{20} \cdot100

\text{Growth percentage}=0.5 \cdot100=50

We can see that insect population is growing at rate of 50% per week.

Now let us write an exponential function for our population.

P(w)=20(1+0.50)^{w}, where P(w) represents population at the end of w weeks.

Let us substitute w=11 in our function to find insect population at the end of 11 weeks.

P(11)=20(1+0.50)^{11}

P(11)=20(1.50)^{11}

P(11)=20\cdot 86.49755859375

P(11)=1729.951171875\approx 1730

Therefore, population of insects at the end of 11th week will be 1730.  



7 0
3 years ago
Help me to get 50 points
trasher [3.6K]

the ansewer is the third one

7 0
2 years ago
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