Match each set of points with the quadratic function whose graph passes through those points. f(x) = x2 − 2x − 15 f(x) = -x2 − 2x + 15 f(x) = -x2 + 2x − 15 (0,-15), (1,-14), (2,-15) (-2,15), (-1,16), (0,15) (-3,0), (0,-15), (5,0)

f(x) = x² - 2x - 15 passes through (-3 , 0) , (0 , -15) , (5 , 0)

f(x) = -x² - 2x + 15 passes through (-2 , 15) , (-1 , 16) , (0 , 15)

f(x) = -x² + 2x - 15 passes through (0 , -15) , (1 , -14) , (2 , -15)

Step-by-step explanation:

* Lets explain how to solve this question

- To find the points whose graph passes through them substitute the

  x-coordinate in the function if the answer is the same with the

  y-coordinate of the point then the graph passes through this point

  lets do that

- Check the first set of points with the first function

# Pint (0 , -15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (0)² - 2(0) - 15 = -15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , -15)

# Pint (1 , -14)

∵ f(x) = x² - 2x - 15

∴ f(0) = (1)² - 2(1) - 15 = -16 ⇒ not same value of y-coordinate

∴ The graph of the function does not pass through point (1 , -14)

∴ The graph does not pass through this set of points

- Check the second set of points with the first function

# Pint (-2 , 15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (-2)² - 2(-2) - 15 = 4 + 4 - 15 -7 ⇒ not same value of y-coordinate

∴ The graph of the function does not pass through point (-2 , 15)

∴ The graph does not pass through this set of points

- Check the third set of points with the first function

# Pint (-3 , 0)

∵ f(x) = x² + 2x - 15

∴ f(0) = (-3)² - 2(-3) - 15 = 9 + 6  -15 = 0 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-3 , 0)

# Pint (0 , -15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (0)² - 2(0) - 15 = -15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , -15)

# Pint (5 , 0)

∵ f(x) = x² + 2x - 15

∴ f(0) = (5)² - 2(5) - 15 = 25 - 10  -15 = 0 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (5 , 0)

∴ The graph passes through this set of points

* f(x) = x² - 2x - 15 passes through (-3 , 0) , (0 , -15) , (5 , 0)

- Check the first set of points with the second function

# Pint (0 , -15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(0)² - 2(0) + 15 = 15 ⇒ not same value of y-coordinate

∴ The graph of the function does not passes through point (0 , -15)

∴ The graph does not pass through this set of points

- Check the second set of points with the second function

# Pint (-2 , 15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(-2)² - 2(-2) + 15 = -4 + 4 + 15 = 15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-2 , 15)

# Pint (-1 , 16)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(-1)² - 2(-1) + 15 = -1 + 2 + 15 = 16 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-1 , 16)

# Pint (0 , 15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(0)² - 2(0) + 15 = 15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , 15)

∴ The graph passes through this set of points

* f(x) = -x² - 2x + 15 passes through (-2 , 15) , (-1 , 16) , (0 , 15)

- Now we have the first set of points and the third function

∴ The graph passes through this set of points

∴ f(x) = -x² + 2x - 15 passes through (0 , -15) , (1 , -14) , (2 , -15)

can someone answer this question plz

step-by-step explanation:

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47