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Len [333]
3 years ago
12

What is the point of intersection when the system of equations below is graphed on the coordinate plane?

Mathematics
1 answer:
WARRIOR [948]3 years ago
8 0

Neither of the options are correct for the given equations.

Step-by-step explanation:

-x+y=4...................................................1

16x+y=-3...............................................2

Subtract equation 1 and 2

                              -x+y- 16x -y = 4+3

                               ⇒ -17x = 7

                               ⇒x= \frac{7}{-17\\}

                              ⇒x= -0.4117

so,

                        16×-0.4417 + y = -3

                       ⇒y= -3+6.5872

                      ⇒y= 3.5872

So the point of intersection is (-0.4117, 3.5872)

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PLEASE PLEASE HELP ​
nekit [7.7K]

Answer:68.3 degrees

Step-by-step explanation:

The diagram of the triangle ABC is shown in the attached photo. We would determine the length of side AB. It is equal to a. We would apply the cosine rule which is expressed as follows

c^2 = a^2 + b^2 - 2abCos C

Looking at the triangle,

b = 75 miles

a = 80 miles.

Angle ACB = 180 - 42 = 138 degrees. Therefore

c^2 = 80^2 + 75^2 - 2 × 80 × 75Cos 138

c^2 = 6400 + 5625 - 12000Cos 138

c^2 = 6400 + 5625 - 12000 × -0.7431

c^2 = 12025 + 8917.2

c = √20942.2 = 144.7

To determine A, we will apply sine rule

a/SinA = b/SinB = c/SinC. Therefore,

80/SinA = 144.7/Sin 138

80Sin 138 = 144.7 SinA

SinA = 53.528/144.7 = 0.3699

A = 21.7 degrees

Therefore, theta = 90 - 21.7

= 68.3 degees

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Step-by-step explanation:
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