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il63 [147K]
3 years ago
13

The product of three integers x, x+2, and x-5 is 240. What are the integers?

Mathematics
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

  8, 10, 3

Step-by-step explanation:

For finding real solutions to cubic equations and those of higher degree, I find a graphing calculator to be useful. The problem can be cast in the form ...

  f(x) = 0

where f(x) is the difference between the product of the integers and 240:

  f(x) = x(x +2)(x -5) -240

Then all we need to do is ask the calculator to show the x-intercept. It is x=8.

So, the three integers are ...

  • x = 8
  • x+2 = 10
  • x -5 = 3

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Which ordered pair is the solution to the system of equations?
SOVA2 [1]
To find the solution, we use the substitution method. 

x+y=-1
<span>x-3y=11
</span>
x+y = -1-y
  -y       

x = -1-y

Now apply the value of x into the other equation.

x-3y=11
-1-y-3y = 11

Combine like terms

-4y -1 = 11
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Now, apply the value of Y to one equation to find x.

y = -3

x -3 = -1
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Now we have the value for both, x and y.
x = 2
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Final answer: A. <span>(2, −3)</span>
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1. -28 = -3n + 5<br> what is n??<br> 2.5−n3=−4
STatiana [176]

Answer:

1.

n=11

2.

n =  \sqrt[3]{9}

Step-by-step explanation:

Question 1:

The given equation is:

- 28 =  - 3n + 5

Subtract 5 from both sides:

- 28 - 5 = - 3n

- 33 =  - 3n

Divide boy sides by:

-3

n =  \frac{ - 33}{ - 3}  = 11

Question 2:

The given expresion is :

5 -  {n}^{3}  =  -  4

Subtract 5 from both sides

-  {n}^{3}=  - 4 - 5

This implies that;

-  {n}^{3}=  - 9

{n}^{3}=  9

n =  \sqrt[3]{9}

5 0
3 years ago
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