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m_a_m_a [10]
3 years ago
6

An athlete is tracking his weight

Mathematics
2 answers:
ycow [4]3 years ago
7 0

Answer:

His rate of change is 7 pounds per week.

Step-by-step explanation:

Rate of change = change in weight ÷ change in time

change in weight = 188 - 167 = 21 pounds

change in time = 6 - 3 = 3 weeks

Rate of change = 21 ÷ 3 = 7 pounds/week

blsea [12.9K]3 years ago
7 0

Answer:

He loses 7 pounds per week

Step-by-step explanation:

After 3 weeks he weighs 188 pounds.

After 6 weeks he weighs 167 pounds.

Weight loss = (188 - 167) pounds = 21 pounds

Time interval = 6 weeks - 3 weeks = 3 weeks

Rate of change = Weight loss ÷ time interval = \frac{21}{3} pounds per week = 7 pounds loss per week.

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use your head

Step-by-step explanation:

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You are given f(0) = 1 and g(0) = -2 and g(1) = 3. What is <br> g ∘ f ( 0 ) <br> equal to?
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Answer:

  • <u><em>g∘f (0) = 1</em></u>

Explanation:

The <em>composition</em> of the <em>functions</em> f and g represented by g ∘ f ( 0 ) means that g is applied to f(0), i.e f(0) is the input to the function g.

Since f(0) = 1, you are going fo find g(1):

         \( g\circ f(x)=g(f(x)) \)\\\\\( g\circ f(0)=g(f(0)) \)\\\\\( g\circ f(0)=g(1) \)=3

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Compare and contrast a rhombus to a square. Make sure to address items such as sides and angles HELP PLZ!!!
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Prime factor
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Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

A_0 = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
3 years ago
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