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Sliva [168]
4 years ago
6

The population of a town with an initial population

Mathematics
1 answer:
Maslowich4 years ago
4 0

Answer:

5 years = 89,940

10 years = 107, 856

Step-by-step explanation:

75000 * 1.037 ^ 5 for the 5 years one

Then do the answer (89940) and times by 1.037 ^ 5. For 10 years

You do it to the power of 5 because you are adding on another 5 years

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How to solve <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B4%7D%20x%20%7B%7D%5E%7B2%7D%20%20%3D%20x%20%2B%202" id="TexFormu
Vikentia [17]

Answer:

➩ x=2,-\frac{2}{3}

Step-by-step explanation:

\sqrt{4x^2}=x+2

➨ We can also solve by completing both squares, however. Since we can pull out the square root.

\sqrt{4x^2}=|2x| ➩ Define of Absolute Value/Square Root

➩ \sqrt{x^2}=|x|

Thus, our new equation is ➩ |2x|=x+2

To solve an absolute-value equation, let there be two conditions.

➨ Where x ≥ 0

2x=x+2\\

Move x to another side

2x-x=2\\x=2

➨ Where x < 0

-2x=x+2\\-2x-x=2\\-3x=2\\x=\frac{2}{-3}\\x=-\frac{2}{3}

3 0
3 years ago
-6(y+15)=-3y+6<br> what value of y makes the equation true
AlladinOne [14]

Answer:

y = -32

Step-by-step explanation:

-6(y+15)=-3y+6

Distribute

-6y - 90 = -3y +6

Add 6y to each side

-6y -90+6y = -3y+6y +6

-90 = 3y+6

Subtract 6 from each side

-90 -6 = 3y +6-6

-96 = 3y

Divide by 3

-96/3 = 3y/3

-32 = y

7 0
3 years ago
Read 2 more answers
I NEED HELP AS ASAP 4 THE FIRST AND SECOND ONE<br> !!!!
irga5000 [103]
Explaination: when graphing a relation, the set of second elements will be the y-values of the graph.
Range: [-4,-3,-2,0,3,5]
8 0
3 years ago
Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

-6=b

Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

7 0
3 years ago
Jordan invests £700 into his bank account.
bearhunter [10]

Answer:

Jordan will have $ 910.00 after 3 years

Step-by-step explanation

6 0
3 years ago
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