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mario62 [17]
3 years ago
10

FInd the equation of a perpendicular line to y=-2 that passes through the points (4,-2)

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
3 0

Answer:

x = 4

Step-by-step explanation:

y=-2 is a horizontal line so x must be a vertical line through (4,-2)

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Which of the following inequalities matches the graph? graph of an inequality with a dashed vertical line through the point (2,
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The answer is x > 2, let me know if you need why
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Step-by-step explanation:

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3 years ago
Triangle ABC is similar to triangle XYZ by the AA Similarity Postulate. Also, m∠A = 44° and m∠C = 31°. What is m∠Y
Pepsi [2]

Answer

m∠Y = 105 degrees

Step by step explanation

Here ΔABC and ΔXYZ are similar. Therefore, the angles of the two triangles are equal in measure.

m∠A = 44 = m∠X, m∠C = 31 = m∠Z

m∠Y = m∠B

m∠X + m∠Y +m∠Z = 180         {sum of the interior angles add upto 180]

44 + m∠Y + 31 = 180

75 + m∠Y = 180

m∠Y = 180 - 75

m∠Y = 105

Thank you.

3 0
3 years ago
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PLEASE HELP I WILL GIVE BRAINLIEST
krok68 [10]

9514 1404 393

Answer:

  48

Step-by-step explanation:

Evaluate each of the functions by putting the number where x is.

  g(x)=x^2-4\\\\g(-7)=(-7)^2-4=49-4\\\\\boxed{g(-7)=45}\\\\f(x)=x+4\\\\f(-7)=-7+4\\\\\boxed{f(-7)=-3}

Then the difference is ...

  g(-7) -f(-7) = 45 -(-3) = 45 +3 = 48

5 0
3 years ago
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WORTH 100 POINTS!
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Answer:

Step-by-step explanation:

4. h(x) = 2x^2 + 14x - 60

Step-by-step explanation:

Given that h(x) is a quadratic.

Also, h(3) = h(-10) = 0

(A) h(x) = x^2 - 13x - 30

=> h(3) = 3^2 - 13(3) - 30

=> h(3) = 9 - 39 - 30

=> h(3) = -30 - 30

=> h(3) = -60

=> h(3) ≠ 0

(B) h(x) = x^2 - 7x - 30

=> h(3) = 3^2 - 7(3) - 30

=> h(3) = 9 - 21 - 30

=> h(3) = -12 - 30

=> h(3) = -42

=> h(3) ≠ 0

(C) h(x) = 2x^2 + 26x - 60

=> h(3) = 2(3^2) + 26(3) - 60

=> h(3) = 2(9) + 78 - 60

=> h(3) = 18 + 78 - 60

=> h(3) = 96 - 60

=> h(3) = 36

=> h(3) ≠ 0

(D) h(x) = 2x^2 + 14x - 60

=> h(3) = 2(3^2) + 14(3) - 60

=> h(3) = 2(9) + 42 - 60

=> h(3) = 18 + 42 - 60

=> h(3) = 60 - 60

=> h(3) = 0

And

=> h(-10) = 2(-10)^2 + 14(-10) - 60

=> h(-10) = 2(100) - 140 - 60

=> h(-10) = 200 - 200

=> h(-10) = 0

Clearly we have,

=> h(3) = h(-10) = 0

Hence, the correct option is (D) h(x) = 2x^2 + 14x - 60

4 0
3 years ago
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