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7nadin3 [17]
1 year ago
13

A wire 22 cm long has a mass of 374 g. What is the mass of 13 cm of this wire?

Mathematics
1 answer:
taurus [48]1 year ago
7 0

Answer:

<u>2</u><u>2</u><u>1</u><u> </u><u>g</u>

The up pinned pic is of inverse variation typed answer.. If u want word problem type answer here are the steps (EVEN IF THE STEPS ARE DIFFERENT ANSWERS REMAIN SAME)

Step-by-step explanation:

Mass of wire from 22cm = 374g

Mass of wire from 1 cm = 374÷22 = 17g

Mass of wire from 13 cm = 13×17 = <u>2</u><u>2</u><u>1</u><u> </u><u>g</u>

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The quotient of 38 times a number and -4
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Answer:

140y

Step-by-step explanation:

35 times a number times -4 is 35×y×(-4) which is -140y

4 0
1 year ago
Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.Tiles
Ugo [173]

By definition, the average rate of change is given by:

AVR = \frac{f(x2)-f(x1)}{x2-x1}

We evaluate each of the functions in the given interval.

We have then:

For f (x) = x ^ 2 + 3x:

Evaluating for x = -2:

f (-2) = (-2) ^ 2 + 3 (-2)\\f (-2) = 4 - 6\\f (-2) = - 2

Evaluating for x = 3:

f (3) = (3) ^ 2 + 3 (3)\\f (3) = 9 + 9\\f (3) = 18

Then, the AVR is:

AVR = \frac{18-(-2)}{3-(-2)}

AVR = \frac{18+2}{3+2}

AVR = \frac{20}{5}

AVR = 4


For f (x) = 3x - 8:

Evaluating for x =4:

f (4) = 3 (4) - 8\\f (4) = 12 - 8\\f (4) = 4

Evaluating for x = 5:

f (5) = 3 (5) - 8\\f (5) = 15 - 8\\f (5) = 7

Then, the AVR is:

AVR = \frac{7-4}{5-4}

AVR = \frac{3}{1}

AVR = 3


For f (x) = x ^ 2 - 2x:

Evaluating for x = -3:

f (-3) = (-3) ^ 2 - 2 (-3)\\f (-3) = 9 + 6\\f (-3) = 15

Evaluating for x = 4:

f (4) = (4) ^ 2 - 2 (4)\\f (4) = 16 - 8\\f (4) = 8

Then, the AVR is:

AVR = \frac{8-15}{4-(-3)}

AVR = \frac{-7}{4+3}

AVR = \frac{-7}{7}

AVR = -1


For f (x) = x ^ 2 - 5:

Evaluating for x = -1:

f (-1) = (-1) ^ 2 - 5\\f (-1) = 1 - 5\\f (-1) = - 4

Evaluating for x = 1:

f (1) = (1) ^ 2 - 5\\f (1) = 1 - 5\\f (1) = - 4

Then, the AVR is:

AVR = \frac{-4-(-4)}{1-(-1)}

AVR = \frac{-4+4}{1+1}

AVR = \frac{0}{2}

AVR = 0


Answer:

from the greatest to the least value based on the average rate of change in the specified interval:


f(x) = x^2 + 3x interval: [-2, 3]

f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]


4 0
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1 year ago
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Answer:

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Step-by-step explanation:

for 120 student they would be 4 teacher needed per 30 student

4:120

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Step-by-step explanation: hope this helps

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