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erastova [34]
3 years ago
12

A line has a slope of 2/3 and a y-intercept of −5. Which equations represent the line? Choose all answers that are correct. (MUL

TIPLE CHOICE)
a. -2x + 3y = -15
b. y= 2/3x - 5
c. 6y = 4x - 30
d. (y + 3) = 2/3 (x - 3)
Mathematics
1 answer:
marin [14]3 years ago
7 0
A.         -2x + 3y = -15
    -2x + 2x + 3y = 2x - 15
                     3y = 2x - 15
                      3          3
                       y = ²/₃ - 5
A is correct.

B. y = ²/₃ - 5
B is also correct.

C. 6y = 4x - 30
     6           6
       y = ²/₃x - 5
C is also correct.

D. y + 3 = ²/₃(x - 3)
     y + 3 = ²/₃(x) - ²/₃(3)
     y + 3 = ²/₃x - 2
         - 3          - 3
           y = ²/₃x - 5
D is also correct.

The answer to the problem is A, B, C, and D.
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ABCD is a trapezium AD and BC are parallel
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Part (a)

<h3>Answer:  12.1</h3>

-----------------------

Work Shown:

We'll apply the sine rule since we have a known opposite side of AB = 10 and an unknown hypotenuse we want to find BD.

Focus on triangle ABD

sin(angle) = opposite/hypotenuse

sin(D) = AB/BD

sin(56) = 10/x

x*sin(56) = 10

x = 10/sin(56)

x = 12.062179

x = 12.1

Make sure your calculator is in degree mode.

===================================================

Part (b)

<h3>Answer:  15.1</h3>

-----------------------

Work Shown:

Draw an xy coordinate grid.

Place point A at the origin (0,0).

Point B is 10 units above this, so B is at (0,10).

Point C is at (18,10) since we move 18 units to the right of B.

Point D is at approximately (6.745085, 0). The 6.745085 is from solving tan(56) = 10/x for x.

Refer to the diagram below.

Apply the distance formula for the points C and D.

C = (x_1,y_1) = (18,10)\\\\D = (x_2,y_2) \approx (6.745085,10)\\\\

d = \text{ distance from C to D}\\\\d = \sqrt{ (x_1-x_2)^2+(y_1-y_2)^2}\\\\d \approx \sqrt{ (18-6.745085)^2+(10-0)^2}\\\\d \approx \sqrt{ 226.673112 }\\\\d \approx 15.055667\\\\d \approx 15.1\\\\

Segment CD is roughly 15.1 cm long.

3 0
3 years ago
Read 2 more answers
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