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zloy xaker [14]
3 years ago
9

What is the average rate of change for this quadratic function for the interval from x =3 to x = 5

Mathematics
1 answer:
Mrrafil [7]3 years ago
5 0

Answer:

A

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

\frac{f(b)-f(a)}{b-a}

here [ a, b [ = [ 3, 5 ]

From the table

f(b) = f(5) = - 26

f(a) = f(3) = - 10

Hence average rate of change

= \frac{-26-(-10)}{5-3} = \frac{-16}{2} = - 8 → A

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Using Matrices to find a co-linear missing point
baherus [9]

Answer:

subject?

Step-by-step explanation:

5 0
2 years ago
John bought 2 5-pound bags of candy to share with his classmates. If there are 25 students in johns class, how much candy will e
Paha777 [63]
Hello,

If you have 2 5-pound bags of candy, that means you have 2 * 5 = 10 pounds of candy to share with the class.

As there are 25 students to share the candy with, divide 10 / 25 = 2/5 of a pound

Each student will receive 2/5 of a pound of candy.

Hope this helps!

8 0
3 years ago
Find the arc measure ML.
Nonamiya [84]
Arc ML = 2×∠MNL
  = 2×43°
  = 86°

_____
The measure of an inscribed angle is half the measure of the arc it intercepts. This simple relationship is used in many geometry problems, so is worth remembering (or at least, writing down in a handy place).
6 0
3 years ago
If C(x) is the cost of producing x units of a commodity, then the average cost per unit is c(x) = C(x)/x. Consider the cost func
nadya68 [22]

Answer:

a) C(1,000)=288,491.11 $

b) c(1,000)=288.49 $/u

c) dC/dx(1,000)=349.74 $/u

d) x=100 u

e) c=220 $/u

Step-by-step explanation:

(a) Find the total cost at a production level of 1000 units.

C(x) = 2,000 + 160x + 4x^{3/2}\\\\C(1,000)=2,000 + 160(1,000) + 4(1,000)^{3/2}\\\\C(1,000)=2,000+160,000+126,491.11= 288,491.11

(b) Find the average cost at a production level of 1000 units.

c(x)=C(x)/x\\\\c(1,000)=C(1,000)/1,000=288,491.11/1,000=288.49

(c) Find the marginal cost at a production level of 1000 units.

\frac{dC}{dx} =0+160+4*(3/2)x^{3/2-1}=160+6x^{1/2}\\\\dC/dx|_{1,000}=160+6*(1,000)^{1/2}=160+189.74=349.74

(d) Find the production level that will minimize the average cost.

c=2,000x^{-1}+160+4x^{1/2}\\\\dc/dx=2,000(-1)x^{-2}+0+4(1/2)x^{-1/2}=0\\\\-2,000x^{-2}+2x^{-1/2}=0\\\\2x^{-1/2}=2,000x^{-2}\\\\x^{-1/2+2}=1,000\\\\x^{3/2}=1,000\\\\x=1,000^{2/3}=100

(e) What is the minimum average cost?

c(x) = 2,000/x + 160 + 4x^{1/2}\\\\C(100)=2,000/100 + 160 + 4(100)^{1/2}=20+160+40=220

5 0
3 years ago
If (3). write the following vectors in column vector form:
Art [367]

Answer:

see explanation

Step-by-step explanation:

Given

a = \left[\begin{array}{ccc}3\\2\\\end{array}\right]

To obtain -3a multiply each of the elements of a by -3

3a = \left[\begin{array}{ccc}-3(3)\\-3(2)\\\end{array}\right] = \left[\begin{array}{ccc}-9\\-6\\\end{array}\right]

To obtain 1.5a multiply each element by 1.5

1.5a = \left[\begin{array}{ccc}1.5(3)\\1.5(2)\\\end{array}\right] = \left[\begin{array}{ccc}4.5\\3\\\end{array}\right]

8 0
3 years ago
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