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Paha777 [63]
3 years ago
8

If AX) = x2 + 2x + 3, what is the average rate of change of AX) over the interval (-4, 6]? OA 51 B. 40 OC. 31 OD 20 OE 4​

Mathematics
1 answer:
Yuki888 [10]3 years ago
4 0

Answer:

E

Step-by-step explanation:

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

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

Here [ a, b ] = [ - 4, 6 ] , then

f(b) = f(6) = 6² + 2(6) + 3 = 36 + 12 + 3 = 51

f(a) = f(- 4) = (- 4)² + 2(- 4) + 3 = 16 - 8 + 3 = 11 , thus

average rate of change = \frac{51-11}{6-(-4)} = \frac{40}{10} = 4 → E

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Veseljchak [2.6K]

Answer:true

Step-by-step explanation:

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Find the distance between the points A (13, 2) and B (7, 10).
Aliun [14]

Answer:

10

Step-by-step explanation:

d = 10

For:

(X1, Y1) = (13, 2)

(X2, Y2) = (7, 10)

Distance Equation Solution:

d=(7−13)2+(10−2)2−−−−−−−−−−−−−−−−−√

d=(−6)2+(8)2−−−−−−−−−−√

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4 0
3 years ago
Find the volume and surface area of this prism.
mina [271]

Answer:

V = 408 cm cubed

SA = 558 cm squared

Step-by-step explanation:

To find the volume of a prism, multiply the area of the base by the height. This is 1/2 times width times height times length.

V =1/2 l*w*h =1/2* 6*8*17 = 408

To find the surface area of a prism, find the area of the triangular base and the area of each rectangular side.

Area of the base is A = 1/2 * b*h = 1/2 * 6 * 8 = 24. Since there are 2 bases, the area is 48.

Area of the rectangular side is A = b*h = 17*10 = 170. Since there are three, the area is 3*170 = 510.

The surface area of the prism is 48 + 510 = 558.

8 0
3 years ago
Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro
Ugo [173]

Answer:

Side\ B = 6.0

\alpha = 56.3

\theta = 93.7

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with \alpha, \beta, \theta

[See Attachment for Triangle]

A = 10cm

C = 12cm

\beta = 30

What the question is to calculate the third length (Side B) and the other 2 angles (\alpha\ and\ \theta)

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

B^2 = A^2 + C^2 - 2ABCos\beta

Substitute: A = 10,  C =12; \beta = 30

B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30

B^2 = 100 + 144 - 240*0.86602540378

B^2 = 100 + 144 - 207.846096907

B^2 = 36.153903093

Take Square root of both sides

\sqrt{B^2} = \sqrt{36.153903093}

B = \sqrt{36.153903093}

B = 6.0128115797

B = 6.0 <em>(Approximated)</em>

Calculating Angle \alpha

A^2 = B^2 + C^2 - 2BCCos\alpha

Substitute: A = 10,  C =12; B = 6

10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 180 - 144 *Cos\alpha

Subtract 180 from both sides

100 - 180 = 180 - 180 - 144 *Cos\alpha

-80 = - 144 *Cos\alpha

Divide both sides by -144

\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}

\frac{-80}{-144} = Cos\alpha

0.5555556 = Cos\alpha

Take arccos of both sides

Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)

Cos^{-1}(0.5555556) = \alpha

56.25098078 = \alpha

\alpha = 56.3 <em>(Approximated)</em>

Calculating \theta

Sum of angles in a triangle = 180

Hence;

\alpha + \beta + \theta = 180

30 + 56.3 + \theta = 180

86.3 + \theta = 180

Make \theta the subject of formula

\theta = 180 - 86.3

\theta = 93.7

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