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Lesechka [4]
3 years ago
13

5(x+1)=20 What is is the answer?

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
3 0

Steps to solve:

5(x + 1) = 20

~Distributive property

5x + 5 = 20

~Subtract 5 to both sides

5x + 5 - 5 = 20 - 5

~Simplify

5x = 15

~Divide 5 to both sides

5x/5 = 15/5

~Simplify

x = 3

Best of Luck!

seropon [69]3 years ago
3 0

Answer: x = 3

Step-by-step explanation: To solve this equation for <em>x,</em> our first step is to distribute the 5 through the parentheses on the left side.

5 times <em>x</em> is 5x and 5 times 1 is 5.

So we have 5x + 5 = 20.

Next, we need to isolate the <em>x</em> term by subtracting 5 from both sides of the equation to get 5x = 15.

Finally, we divide both sides of the equation by 5 to get <em>x = 3 </em>which is the solution to our equation.

It's very easy to make careless mistakes when using the distributive property so make sure to check your answer.

In this problem, to check our answer, we substitute 3 back in for <em>x</em> in the original equation and we have 3 + 1 which is 4 and 5 x 4 which is 20 so our answer checks.

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A population of squirrels is growing in a Louisiana forest with a monthly growth constant of 55 percent. If the initial count is
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The population of squirrels in the Louisiana forest growing monthly at a rate of 5% currently from 100, will be <u>182</u> after a year.

The final value of any quantity growing constantly at a particular rate is given as V = V_{0}e^{rt} ,

where V is the final value, V₀ is the initial value, r is the rate of growth per time period, and t is the number of time periods.

The current population of squirrels (V₀) = 100.

The growth rate (r) = 5% per month.

The time period (t) = 1 year = 12 months.

Hence, the final population of squirrels (V), is given as:

V = V_{0}e^{rt} ,

or, V = 100e^{(0.05*12)} ,

or, V = 100e^{0.60} ,

or, V = 100*1.822119,

or, V = 182.2119 ≈ 182.

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2 years ago
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If x2=18x+y and y2=18y+x, then find the value of root
mezya [45]

Answer:

\sqrt{x^2+y^2+1}=18

Step-by-step explanation:

We are given:

x^2=18x+y\qquad\qquad[1]

y^2=18y+x\qquad\qquad[2]

Subtracting [1] and [2]:

x^2-y^2=18x+y-(18y+x)

Operating:

x^2-y^2=18x+y-18y-x

Recall:

x^2-y^2=(x-y)(x+y)

Substituting:

(x-y)(x+y)=18x+y-18y-x

Rearranging:

(x-y)(x+y)=18x-18y-(x-y)

(x-y)(x+y)=18(x-y)-(x-y)

Dividing by x-y (recall x≠y):

x+y=18-1=17

x+y=17\qquad[3]

Now we add [1] and [2]:

x^2+y^2=18x+y+18y+x

Rearranging:

x^2+y^2=18x+18y+x+y

x^2+y^2=18(x+y)+(x+y)

x^2+y^2=19(x+y)

Substituting from [3]

x^2+y^2=19*17=323

Adding 1:

x^2+y^2+1=324

Taking the square root:

\sqrt{x^2+y^2+1}=\sqrt{324}=18

Thus:

\mathbf{\sqrt{x^2+y^2+1}=18}

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