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8_murik_8 [283]
2 years ago
6

Please Help!! 15 points!!!! Create a real world situation for -8 + (-12) and solve

Mathematics
1 answer:
Vanyuwa [196]2 years ago
8 0

Answer:

-20

Step-by-step explanation:

-8-12 = -20

i can't think of a real life solution but i hope this helps a bit.

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Please help me with this problem. I have tried many times and have gotten it wrong. I need help especially on the last part. Can
Arlecino [84]

1) The domain is all the possible x values in the function so it would be [-4,4]

2) There are only 3 zeros shown on the graph and they are (-2, 0) (0, 0) (2, 0) the zeros are the value of x when y = 0.

3)The function is positive/Negative is asking for what x values make the y values positive aka interval notation. The function is positive if x = (0, 2) because 0 and 2 aren't included you use parentheses () instead of brackets []  

The function is negative if x [-4, -2), (-2,0), (2,4]

5 0
3 years ago
Estimate 4281+7028 by first rounding each number to the nearest thousand.
Readme [11.4K]
4000+7000= 11,000 nearest thousands
4281+7028= 11,309 actual number
7 0
2 years ago
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The equation of a parabola is y=x^2-10x+27. Write the equation in vertex form
azamat

Answer:

\large \boxed{y = (x - 5)^{2} + 2 }

Step-by-step explanation:

y = x² - 10x + 27

y = ax² + bx + c

This is the general form of the equation for a parabola.

We must convert it to the vertex form

y = (x - h)² + k, where (h,k) are the coordinates of the vertex.

We can do this by completing the square.

\begin{array}{rcll}y & = & x^{2} - 10x + 27 & \\y - 27 & = & x^{2} - 10x & \text{Subtracted 27 from each side}\\y - 27&= & x^{2} - 10x + 25 - 25 & \text{Added and subtracted (b/2)}^{2}\\y - 27&= & (x - 5)^{2} - 25 & \text{Wrote the first three terms as the square of a binomial}\\y& = & \mathbf{(x - 5)^{2} + 2} & \text{Added 27 to each side}\\\end{array}\\\text{The vertex form of the parabola is $\large \boxed{\mathbf{y = (x - 5)^{2} + 2 }}$}The figure below shows that your parabola has its vertex at (5,2).

4 0
3 years ago
2m^2n^4/6m^5n^3 assume no variable equals zero.
Andrews [41]
<span>2m^2n^4/6m^5n^3
           n
 =   --------
         3m^3</span>
8 0
3 years ago
What is the value of x in the equation 4x + 8y = 40, when y = 0.8?
storchak [24]
To find your answer you must substitute y for 0.8
4x + (8)(0.8) = 40
Now solve for x
====================================================================
SOLVING...

<span>4x + (8)(0.8) = 40
</span><span>4x + 6.4 </span>= <span>40
</span>
Subtract 6.4 from each side
<span><span><span>4x</span>+ 6.4 </span>− 6.4 </span>= <span>40 − <span>6.4
</span></span>4x = 33.6

Divide each side by 4
4x ÷ 4 = 33.6 ÷ 4
x = 8.4
3 0
3 years ago
Read 2 more answers
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