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3 years ago

Which fine fits the data graphed below?

Mathematics

2 answers:

pychu [463]3 years ago

8 0

the answer issssssssssss A

pshichka [43]3 years ago

4 0

Answer is A ................

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Calculate the average rate of change over the interval 2

Morgarella [4.7K]

Answer:

Average rate of change over the interval 2<= x <= 5:

y = 3x + 5: 3

y = 3x^2 + 1: 21

y = 3^x: 78

Step-by-step explanation:

2<= x <= 5

Average rate of change over the interval 2<= x <= 5:

y = 3x + 5

y(5) = 3(5) + 5 = 20

y(2) = 3(2) + 5 = 11

Average rate of change = (20 - 11)/(5-2) = 9/3 = 3

y = 3x^2 + 1

y(5) = 3(5^2) + 1 = 75 + 1 = 76

y(2) = 3(2^2) + 1= 13

Average rate of change = (76 - 13)/(5-2) = 63/3 = 21

y = 3^x

y(5) = 3^5 = 243

y(2) = 3^2 =9

Average rate of change = (243-9)/(5-2) = 234/3 = 78

8 0

2 years ago

There are 360 mins for a monthly cell phone minutes for 4 people in the family, can each person get the same number of minutes o

telo118 [61]

yes, 360/4 = 90 minutes

6 0

3 years ago

If a person rides their bike 40 miles in 5 hours, then they rode their bike at a speed of ___ Mph?

monitta

Answer:

8 mph

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

In 4 minutes, a conveyor belt moves 800 pounds of recyclable aluminum from the delivery truck to a storage area. A smaller belt

kkurt [141]

Answer:

It takes 2 minutes 4 seconds to move the cans to storage area if both the belts are used.

Step-by-step explanation:

Given - In 4 minutes, a conveyor belt moves 800 pounds of recyclable

aluminum from the delivery truck to a storage area. A smaller belt

moves the same quantity of cans the same distance in 6 minutes.

To find - If both belts are used, find how long it takes to move the cans

to the storage area.

Proof -

Given that,

Larger belt take 4 minutes to transfer the aluminium from delivery truck to storage area.

⇒ Time taken by Larger belt = 4 min/ shifting

Also,

Smaller belt take 6 minutes to transfer the aluminium from delivery truck to storage area.

⇒ Time taken by Smaller belt = 6 min/ shifting

Now,

As we know that Time is inversely proportional to time, so

Rate of Larger Belt = $\frac{1}{4}$ shifting/min

Rate of Smaller Belt = $\frac{1}{6}$ shifting/min

Now,

Let us assume that , if both belts used then,

Total time taken = x min/shifting

Then, Rate = $\frac{1}{x}$ shifting/min

Now,

$\frac{1}{4}$ + $\frac{1}{6}$ = $\frac{1}{x}$

⇒ $\frac{6 + 4}{24} = \frac{1}{x}$

⇒ $\frac{10}{24} = \frac{1}{x}$

⇒10x = 24

⇒x = $\frac{24}{10}$ = 2.4 minutes

∴ we get

It takes 2 minutes 4 seconds to move the cans to storage area if both the belts are used.

3 0

2 years ago

dollars of a Class party is 59 + 13 in where in is the number of people attending what is the cost for 44 people

Alina [70]

Plug in 44
so
59+13(44) = 631

7 0

3 years ago