The function h(x) is quadratic and h(3) = h(–10) = 0. which could represent h(x)?

h(x) = x2 – 13x – 30

h(x) = x2 – 7x – 30

h(x) = 2x2 + 26x – 60

h(x) = 2x2 + 14x – 60

Option D.

Step-by-step explanation:

We can solve this question by finding the values of h(3) and h(-10) from each option given.

Option A.

h(x) = x² - 13x - 30

h(3) = 3² - (13×3) - 30

      = 9 - 39 - 30

      = -60

h(-10) = (-10)² - 13(-10) - 30

         = 100 + 130 - 30

         = 200

Therefore, h(3) ≠ h(-10)

Option B.

h(x) = x² - 7x - 30

h(3) = 3² - 7×3 - 30

      = 9 - 21 - 30

      = -42

h(-10) = (-10)² - 7(-10) - 30

         = 100 + 70 - 30

         = 140

h(3) ≠ h(-10)

Option C.

h(x) = 2x² + 26x - 60

h(3) = 2(3)² + 26×3 - 60

      = 18 + 78 - 60

      = 36

h(-10) = 2(-10)² + 26(-10) - 60

         = 200 - 260 - 60

         = -120

h(3) ≠ h(-10)

Option D.

h(x) = 2x² + 14x - 60

h(3) = 2(3)² + 14(3) - 60

      = 18 + 42 - 60

      = 0

h(-10) = 2(-10)² + 14(-10) - 60

         = 200 - 140 - 60

         = 0

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

Therefore, option D is the answer.

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.