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Marrrta [24]
2 years ago
11

if f is a second degree polynomial function such that f(3) = 5, f' (3) = 6, and f'' (3) = 4, what is the value of f(2)

Mathematics
1 answer:
insens350 [35]2 years ago
6 0

Answer:

that is the true answer..

good luck

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A sequence is defined recursively by f(1)=16 and f(n)= f(n-1)+2n. Find f(4)
lord [1]
Hi,

f(1)=16
f(2)=f(1)+2*2=16+4=20
f(3)=f(2)+2*3=20+6=26
f(4)=f(3)+2*4=26+8=34

8 0
3 years ago
Answer the question below
Lena [83]
To find surface area, you would find the area of each side of the shape. Since this is a cube, each side is a square. So area=side2, So each side is 9, so 9x9=81 for each face. There are 6 faces to a cube, so 6x81=486 in2 is your answer
4 0
3 years ago
Miles incorrectly gave the product of 7 * 2.6 as 14 * 42 to use a place value chart or an area model to help miles understand hi
mrs_skeptik [129]
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7 0
3 years ago
If BE = 2x + 2, BD = 5x – 3, and AE = 4x – 6, what are the values of x and AC?
Keith_Richards [23]

Answer: x=7 and AC = 44 unuts.

Step-by-step explanation:

We know that the diagonals of a parallelogram bisect each other. (i)

Here in parallelogram ABCD , AC and Bd are diagonals intersecting at E.

BE = 2x + 2, BD = 5x – 3, and AE = 4x – 6

Using (i)

BE=\dfrac{BD}{2}\\\\2x+2=\dfrac{5x-3}{2}\\\\ 2(2x+2)=5x-3\\\\ 4x+4= 5x-3\\\\ 5x-4x=4+3\\\\ x= 7

Now , AE = 4(7)-6 = 28-6 = 22

AC =2 AE = 2 (22) =44 units.

Hence, x=7 and AC = 44 unuts.

3 0
3 years ago
Given f'(x) = (1 − x)(4 − x), determine the intervals on which f(x) is increasing or decreasing.
Sophie [7]
The critical points are at x = 1 and x = 4 giving you the intervals (-inf, 1), (1, 4) and (4, inf).

By substituting x values in these 3 intervals, you can see that f'(x) is positive in the first and third intervals and negative in the second interval.

This means that f(x) is increasing in the first and third intervals and decreasing in the second interval.

The answer is D.
8 0
3 years ago
Read 2 more answers
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