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Anna [14]
3 years ago
15

H(-4), h(0), h(4), h(8)

Mathematics
1 answer:
DIA [1.3K]3 years ago
6 0

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

The first two x-values, -4 and 0, fall in the first part of the function definition.

There is no function definition for x = 4, as the second part only applies for x > 4.

The last x-value, 8, falls in the second part of the function definition.

The function is evaluated by putting the x-value where x is in the expression and doing the arithmetic. For example, ...

  h(-4) = -(-4+1)^2 +7 = -(-3)^2 +7 = -9 +7 = -2

  h(0) = 6

  h(4) = undefined

  h(8) = -8/3

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Pls helpppppp anyoneeee
defon

Answer:

16

Step-by-step explanation:

7 0
3 years ago
Help me with this please, these problems are so stressful. Right answers only, 5 stars and a thank you for the right answers!
larisa86 [58]
6 2/3 is the answer I hope this helps

3 0
3 years ago
Please help me with both the questions ty.
Tatiana [17]

Answer:

4) 26+4/842-/×375-^×

5)/^9+37+×+454+12

Step-by-step explanation:

6 0
2 years ago
The fourth term in an arithmetic sequence is 18 and the seventh term is 39. If the first term is a1 , which is an equation for t
evablogger [386]

Answer:

<h2>            aₙ = 7n - 10</h2>

Step-by-step explanation:

The nth term:  a_n=a_1+d(n-1)

So:

    a₄ = a₁ + 3d

18 =  a₁ + 3d   ⇒   3d =  18 - a₁

    a₇ = a₁ + 6d

39 = a₁ + 6d

39 = a₁ + 2(18 - a₁)

39 = a₁ + 36 - 2a₁

39 = 36 - a₁

a₁ = -3

3d = 18 - (-3)

3d = 21

d = 7

So the nth term:

                            a_n = -3 + 7(n-1)\\\\a_n=-3 +7n - 7\\\\a_n = 7n - 10

8 0
2 years ago
Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters = 71 and 2 = 6:25. What per
Alexus [3.1K]

Answer:

2.28%

Step-by-step explanation:

given that the height, in inches, of a 25-year-old man is a normal random variable with parameters = 71 and variance = 6:25.

If X is the height of 25 year old man then X is N(71, 2.5)

a) Probability of men in the 6-footer club are over 6 feet, 5 inches

=P(X>6'5")\\=P(X>77")\\= P(Z>\frac{77-72}{2.5} \\=P(X>2)

= 0.0228

percentage of men in the 6-footer club are over 6 feet, 5 inches

100 times prob

=2.28%

2.28% of men in the 6-footer club are over 6 feet, 5 inches willbe taller than 6 feet and 5 inches

5 0
3 years ago
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