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4 years ago

12

Will adding a negative number to another integer always result in a negative sum? Why?

1 answer:

Ghella [55]4 years ago

7 0

Answer:

D No, the integer with the larger absolute value always determines the sign of the sum.

Step-by-step explanation:

Suppose the integer you start with is +3. Adding -1 or -2 or -3 to that will result in 2, 1, or 0, none of which are negative. Only when you add a negative number with an absolute value greater than 3 will you get a sum that is negative. That is, <em>the number in the sum that has the largest absolute value determines the sign of the result</em>. (This is most important when the signs differ, but it is also true when the signs are the same.)

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3 years ago

Ava wants to draw a parallelogram on the coordinate plane. She

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Answer:

$k =(2,1)$

$JK = 2$

Step-by-step explanation:

Given

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$H = (1,-2)$ --- $(x_2,y_2)$

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See attachment for grid

Solving (a): The coordinates of K

The parallelogram has the following diagonals: IJ and HK

Diagonals bisect one another. So:

Midpoint of IJ = Midpoint of HK

This gives:

$\frac{1}{2}(I + J) = \frac{1}{2}(H+K)$

$\frac{1}{2}(x_3+x_1,y_3+y_1) = \frac{1}{2}(x_2+x,y_2+y)$

$\frac{1}{2}(3+0,-2+1) = \frac{1}{2}(1+x,-2+y)$

$\frac{1}{2}(3,-1) = \frac{1}{2}(1+x,-2+y)$

Multiply through by 2

$(3,-1) = (1+x,-2+y)$

By comparison:

$1 + x = 3$

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Solve for x and y

$x = 3 - 1 =2$

$y = -1 +2 = 1$

So, the coordinates of k is:

$k =(2,1)$

The length of JK is calculated using distance (d) formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

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$k =(2,1)$ ---- $(x_2,y_2)$

So:

$d = \sqrt{(0 - 2)^2 + (1 - 1)^2$

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$d = \sqrt{4$

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Hence:

$JK = 2$

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