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Juli2301 [7.4K]
3 years ago
9

I need help on Find the slope of the line segment with endpoint G(1,3) and H(4,7) without graphing

Mathematics
1 answer:
ololo11 [35]3 years ago
7 0

y2-y1 / x2-x1


7-3 / 4-1

4/3

1 & 1/3 should be correct

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Pleaseee helpppp <br><br> What is the sum of -3 and -12
faltersainse [42]

Answer:

-15

Step-by-step explanation:

When adding together two negatives it works just as it would with positive numbers, just with the negative sign to properly negate it.

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3 years ago
Factor z^2 – 3z – 18. A. (z + 3)(z – 6) B. (z – 3)(z – 6) C. (z + 8)(z – 10) D. (z + 2)(z – 9)
ozzi
Work shown above! Answer would be A!

5 0
3 years ago
PLEASE HELP QUESTION IS WORTH 16 POINTS I WILL GET AN F IF INCORRECT
Alborosie

Answer:

D)

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Step-by-step explanation:

5 0
3 years ago
Please help me out with this!!!!!!!
Bad White [126]

Answer:

A

Step-by-step explanation:

The line on the right has equation y = x + 1

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3 0
3 years ago
A perpendicular bisector, , is drawn through point C on .
Liula [17]

Answer:

The x-intercept of CD is B(18/5,0). The point C(32,-71) lies on the line CD.

Step-by-step explanation:

the x-intercept of CD is[ A(3,0) B(18/5,0) C(9,0) D(45/2,0) ] . Point [ A(-52,117) B(-20,57) C(32,-71) D(-54,-128) ] lies on CD.

Given :

CD is perpendicular bisector of AB.

The coordinates of point A are (-3, 2) and the coordinates of point B are (7, 6).

C is the midpoint of AB.

C=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{7-3}{2},\frac{2+6}{2})=(2,4)

The coordinates of C are (2,4).

Line AB has a slope of:m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{7-(-3)}=\frac{4}{10}=\frac{2}{5}

The product of slopes of two perpendicular lines is -1. Since the line CD is perpendicular to AB, therefore the slope of CD : m_2=-\frac{5}{2}

The point slope form of a line is given by:

y-y_1=m(x-x_1)

The slope of line CD is -\frac{5}{2} and the line passing through the point (2,4), the equation of line CD can be written as:

y-4=-\frac{5}{2}(x-2)\\y=-\frac{5}{2}x+5+4\\y=-\frac{5}{2}x+9       .... (1)

The equation of CD is y=-\frac{5}{2}x+9

In order to find the x-intercept, put y=0.

0=-\frac{5}{2}x+9\\\frac{5}{2}x=9\\x=\frac{18}{5}

Therefore the x-intercept of CD is B(18/5,0).

Put x=-52 in eq(1).

y=-\frac{5}{2}(-52)+9=139

Put x=-20 in eq(1).

y=-\frac{5}{2}(-20)+9=59

Put x=32 in eq(1)

y=-\frac{5}{2}(32)+9=-71

Put x=-54 in eq1).

y=-\frac{5}{2}(-54)+9=144

Thus, only point (32,-71) satisfies the equation of CD. Therefore the point C(32,-71) lies on the line CD.

8 0
4 years ago
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