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3 years ago

14

The cheerleaders are making a banner that is 8 feet wide. The length of the banner is 1 1/3 times the width of the banner. How l

ong is the banner?

1 answer:

vfiekz [6]3 years ago

3 0

Use a calculator :)))

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Two functions f and g are given. Show that the growth rate of the linear function is constant and that the relative growth rate

joja [24]

Answer:

linear function growth rate: 8.5

Step-by-step explanation:

The growth rate of the linear function is the coefficient of t: 8.5. (It is a constant.)

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The growth rate of g(t) is its derivative: g'(t) = (1/8)(160e^(t/8)) = 20e^(t/8). Then the relative growth rate is ...

g'(t)/g(t) = (20e^(t/8))/(160e^(t/8)) = 20/160 = 1/8

It is a constant.

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3 years ago

Which figure represents the final image after composition t3, -1 is applied to rectangle

harkovskaia [24]

Answer:T-3, -1 applied to a rectangle hm

Step-by-step explanation:

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3 years ago

The box of cereal shown below has a volume of 297.16 cubic inches. The width is 4.6 inches and the length is 6.8 inches. In inch

finlep [7]

Answer:

9.5 inches

Step-by-step explanation:

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3 years ago

5-4+7x+1 use the drop down menu

In-s [12.5K]

Answer: 2+7x

Step-by-step explanation:

First you have to do PEMDAS. Then you get this answer.

7 0

3 years ago

Read 2 more answers

The mass, m grams, of a radioactive substance, present at time t days after first being observed, is given by the formula m=24e^

Reika [66]

Answer:

(i) The value of<em> m</em> when t = 30 is 13.2

(ii) The value of <em>t</em> when the mass is half of its value at t=0 is 34.7

(iii) The rate of the mass when t=50 is -0.18

Step-by-step explanation:

(i) The <em>m</em> value when t = 30 is:

$m = 24e^{-0.02t} = 24e^{-0.02*30} = 13.2$

Then, the value of<em> m</em> when t = 30 is 13.2

(ii) The value of the mass when t=0 is:

$m_{0} = 24e^{-0.02t} = 24e^{-0.02*0} = 24$

Now, the value of <em>t </em>is:

$ln(\frac{m_{0}/2}{24}) = -0.02t$

$t = -\frac{ln(\frac{24}{2*24})}{0.02} = 34.7$

Hence, the value of <em>t</em> when the mass is half of its value at t=0 is 34.7

(iii) Finally, the rate at which the mass is decreasing when t=50 is:

$\frac{dm}{dt} = \frac{d}{dt}(24e^{-0.02t}) = 24(e^{-0.02t})*(-0.02) = -0.48* (e^{-0.02*50}) = -0.18$

Therefore, the rate of the mass when t=50 is -0.18.

I hope it helps you!

3 0

2 years ago