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3 years ago

9

What is the average rate of change of the function over the interval x = 0 to x = 6? f(x)=2x−1 3x+5 Enter your answer, as a frac

tion, in the box.

1 answer:

Tatiana [17]3 years ago

8 0

Answer:

-11

Step-by-step explanation:

Our function is 2x-13x+5

2x-13x+5= -11x+5
F(0)= 5 and F(6)= -11*6+5 = -61
let m be the average change : m= (-61-5)/6= -11

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Megan uses 2/3 cups of almonds to make 4 cups of trail mix. Using the same ratio, how many cups of almonds would megan need to m

lakkis [162]

1 and 1/2
9 divided by 4 = 2.25
2.25x2/3=1 1/2

3 0

3 years ago

Read 2 more answers

A car is traveling is at a speed of 99 km/h. what is the car speed in kilometers per minute how many kilometers for the car trav

Ad libitum [116K]

Answer:

1.65 kilometers a minute and you would travel 3.3 kilometers in two minuets

Step-by-step explanation:

Im pretty sure that is the answer because 99/60 = 1.65 and 1.65 * 2 = 3.3

7 0

2 years ago

Cos90degrees cos30 degrees + sec-squared45degrees sin30degrees - cos270degrees sin180degrees

yawa3891 [41]

Step-by-step explanation:

$\cos 90 \cos 30 + \sec^2 45 \sin 30 - \cos 270 \sin 180$

But we know that

$\cos 90 = \cos 270 = \sin 180 = 0$

and

$\sin 30 = \dfrac{\sqrt{3}}{2} \:\:\text{and}\:\: \sec 45 = \sqrt{2}$

so this leaves us with just

$\sec^2 45 \sin 30 = (\sqrt{2})^2 \left(\dfrac{\sqrt{3}}{2} \right)= \sqrt{3}$

6 0

3 years ago

Approximate the change in the volume of a sphere when its radius changes from r = 40 ft to r equals 40.05 ft (Upper V (r )equal

alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere ( $V$ ), measured in cubic feet, is represented by the following formula:

$V = \frac{4\pi}{3}\cdot r^{3}$

Where $r$ is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

$\Delta V = \frac{\partial V}{\partial r}\Delta r$

The derivative of the volume as a function of radius is:

$\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}$

Then, the change in volume is expanded:

$\Delta V = 4\pi \cdot r^{2}\cdot \Delta r$

If $r = 40\,ft$ and $\Delta r = 40\,ft-40.05\,ft = 0.05\,ft$ , the change in the volume of the sphere is approximately:

$\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)$

$\Delta V \approx 1005.310\,ft^{3}$

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

7 0

3 years ago

Suppose that 24% of the students in the first group answered yes and that 73% of the students in the second group answered yes.

Lera25 [3.4K]

Answer:

Price Discrimination OR Law of Demand; according to the complete question.

Step-by-step explanation:

24% of the students in the first group answered yes.

73% of the students in the second group answered yes.

More students in the second group were willing to pay $75 for the pair of jeans BECAUSE they were told that the normal price was much higher.

From this information, I guess that the first group was told (by the jeans vendor probably) that the $75 was higher than the normal price of the jeans. This will be the reason why a lesser percentage of students in Group A are willing to purchase the pair of jeans.

This is an example of PRICE DISCRIMINATION effect on decision making. Price discrimination is used in product marketing.

The same pair of jeans in Situation A cost higher than the normal price while in Situation B it cost lower than the normal price. Even though the figure given is static at $75 in both cases, the data that follows in the question tells it as 2 different prices; one favourable to the buyers and another not so favourable to the buyers.

The LAW OF DEMAND also applies here. The higher the price, the lesser the quantity demanded (by a group of students) and the lower the price, the higher the quantity demanded.

5 0

3 years ago