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3 years ago

Need help asap!!! Will give lots of points!!!

Mathematics

1 answer:

nikdorinn [45]3 years ago

3 0

$\boxed{16x^2}$

$\underline{\textrm{Solve for perimeter}}$

$16x \div 4 = 4x$

$\underline{\textrm{Solve for area}}$

$4x * 4x = 16x^2$

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The missing color is D. white.

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3 years ago

Two intersecting lines will form how many pairs of congruent angles?

HACTEHA [7]

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3 years ago

there are 5 pounds of apples in one bag make a table to show how many pounds of apples are in 6 bags. how do you know your answe

Dmitrij [34]

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Then 6 bags must equal six times the weight of the first bag
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2 years ago

Read 2 more answers

Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.

Romashka [77]

By definition, the average change of rate is given by:
$AVR = \frac{f(x2)-f(x1)}{x2-x1}$
We will calculate AVR for each of the functions.
We have then:

f(x) = x^2 + 3x interval: [-2, 3]:
$f(-2) = x^2 + 3x = (-2)^2 + 3(-2) = 4 - 6 = -2

f(3) = x^2 + 3x = (3)^2 + 3(3) = 9 + 9 = 18$
$AVR = \frac{-2-18}{-2-3}$
$AVR = \frac{-20}{-5}$
$AVR = 4$

f(x) = 3x - 8 interval: [4, 5]:
$f(4) = 3(4) - 8 = 12 - 8 = 4 f(5) = 3(5) - 8 = 15 - 8 = 7$
$AVR = \frac{7-4}{5-4}$
$AVR = \frac{3}{1}$
$AVR = 3$

f(x) = x^2 - 2x interval: [-3, 4]
$f(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15

f(4) = (4)^2 - 2(4) = 16 - 8 = 8$
$AVR = \frac{8-15}{4+3}$
$AVR = \frac{-7}{7}$
$AVR = -1$

f(x) = x^2 - 5 interval: [-1, 1]
$f(-1) = (-1)^2 - 5 = 1 - 5 = -4

f(1) = (1)^2 - 5 = 1 - 5 = -4$
$AVR = \frac{-4+4}{1+1}$
$AVR = \frac{0}{2}$
$AVR = 0$

Answer:
these functions from the greatest to the least value based on the average rate of change are:
f(x) = x^2 + 3x
f(x) = 3x - 8
f(x) = x^2 - 5
f(x) = x^2 - 2x

5 0

3 years ago

H(t)= (t+3)^2 + 5

oee [108]

Answer:

The average rate of change for the given function in the interval (-5, -1) is 0 (zero)

Step-by-step explanation:

The average rate of change of a function over an interval is the quotient between the difference between the function evaluated at the ends of the interval divided by the length of the interval. That is for our case:

the average rte of change of h(t) in the interval (-5, -1) is:

$\frac{h(-1)-h(-5)}{-1+5}$

so we find:

$h(-1)=(-1+3)^2+5=2^2+5=4+5=9\\and\\h(-5)=(-5+3)^2+5=(-2)^2+5=4+5=9$

then the average rate of change becomes:

$\frac{h(-1)-h(-5)}{-1+5}=\frac{9-9}{4} =\frac{0}{4} =0$

5 0

2 years ago