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Mice21 [21]
2 years ago
12

What is the value of -2(8-(-2)) -8(17-12) ? 1 ) -2 02 O4

Mathematics
1 answer:
guajiro [1.7K]2 years ago
5 0

Answer:

C. 2

Step-by-step explanation:

\frac{-8(17-12)}{-2(8-(-2))}

Subtract inside the parenthesis inside the numerator.

\frac{-8(5)}{-2(8-(-2))}

Multiply -8 and 5 in the numerator.

\frac{-40}{-2(8-(-2))}

Rewrite the denominator knowing that x - (-y) = x + y.

\frac{-40}{-2(8+2)}

Add inside the parenthesis in the denominator.

\frac{-40}{-2(10)}

Multiply -2 and 10 in the denominator.

\frac{-40}{-20}

Rewrite the fraction knowing that (-x)/(-y) = x/y.

\frac{40}{20}

Divide 40 by 20 to get: 2.

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Lesechka [4]
H ( x ) = - 6 + x
m = 1 ( the slope )
b = - 6 ( y - intercept )
x - intercept: 
0 = - 6 + x
x = 6 
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Answer:
B ) Quadrant II, because the slope is positive and y-intercept is negative.
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rodikova [14]

Answer:

6b

Step-by-step explanation:

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Find the distance between -4.3 and 0.8 on a number line as a decimal
evablogger [386]

Answer:

5.1

Step-by-step explanation:

Add 4.3 to 0.8

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3 years ago
What monomial multiplied by - 3x ^ 3 * y equals 18x ^ 9 * y ^ 4 Explain how you were able to determine the coefficient and the e
Archy [21]

Answer:

-6x^6y^5

Step-by-step explanation:

you have to find p such that

(-3x^{3}y)(p) = 18x^{9}y^{4}

So just divide to find p

p = \frac{18x^{9}y^{4} }{-3x^{3}y} = -6x^{6}y^{5}

7 0
3 years ago
For each system of equations, drag the true statement about its solution set to the box under the system?
natta225 [31]

Answer:

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

 of x and y are the same and the numerical terms are different

- The system of equation has infinity many solutions if the

   coefficients of x and y are the same and the numerical terms

   are the same

- The system of equation has one solution if at least one of the

  coefficient of x and y are different

* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

  numerical terms

∴ They have zero equation

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

- Substitute the value of x in equation (1) or (2) to find y

∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

3 0
3 years ago
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