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4 years ago

The length of the diagonal of a square is 8 cm, how long are the sides of the square?

Mathematics

1 answer:

Svetach [21]4 years ago

8 0

Answer:

5.66

Step-by-step explanation:

The sides of a square are equal. Let's call the length

Use Pythagoras' Theorem.

Square the sides and add them together....

√

5.66

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Need help with this

kicyunya [14]

The answer is 70%. 28(number of all the sports besides soccer) divided by 40(number of all sports) is 0.7. Multiply 0.7 by 100 and you will get 70%.

8 0

4 years ago

I have this for my math homework but I don’t understand it or how to do it.

rewona [7]

Divide 5/6 by 2/3 by inverting (turning upside down) that 2/3 and then multiplying:

5 3 15 5
--- * --- = ----- = --- (answer).
6 2 12 4

Hope this both helps you and raises some questions. Mind trying the next problem and posting your result? Glad to give you feedback on your work.

6 0

3 years ago

The population of Americans age 55 and older as a percentage of the total population is approximated by the function f(t) = 10.7

MatroZZZ [7]

Answer:

Part A)

About 0.51% per year.

Part B)

About 0.30% per year.

Part C)

About 28.26%.

Step-by-step explanation:

We are given that the population of Americans age 55 and older as a percentange of the total population is approximated by the function:

$f(t) = 10.72(0.9t+10)^{0.3}\text{ where } 0 \leq t \leq 20$

Where t is measured in years with t = 0 being the year 2000.

Part A)

Recall that the rate of change of a function at a point is given by its derivative. Thus, find the derivative of our function:

$\displaystyle f'(t) = \frac{d}{dt} \left[ 10.72\left(0.9t+10\right)^{0.3}\right]$

Rewrite:

$\displaystyle f'(t) = 10.72\frac{d}{dt} \left[(0.9t+10)^{0.3}\right]$

We can use the chain rule. Recall that:

$\displaystyle \frac{d}{dx} [u(v(x))] = u'(v(x)) \cdot v'(x)$

Let:

$\displaystyle u(t) = t^{0.3}\text{ and } v(t) = 0.9t+10 \text{ (so } u(v(t)) = (0.9t+10)^{0.3}\text{)}$

Then from the Power Rule:

$\displaystyle u'(t) = 0.3t^{-0.7}\text{ and } v'(t) = 0.9$

Thus:

$\displaystyle \frac{d}{dt}\left[(0.9t+10)^{0.3}\right]= 0.3(0.9t+10)^{-0.7}\cdot 0.9$

Substitute:

$\displaystyle f'(t) = 10.72\left( 0.3(0.9t+10)^{-0.7}\cdot 0.9 \right)$

And simplify:

$\displaystyle f'(t) = 2.8944(0.9t+10)^{-0.7}$

For 2002, t = 2. Then the rate at which the percentage is changing will be:

$\displaystyle f'(2) = 2.8944(0.9(2)+10)^{-0.7} = 0.5143...\approx 0.51$

Contextually, this means the percentage is increasing by about 0.51% per year.

Part B)

Evaluate f'(t) when t = 17. This yields:

$\displaystyle f'(17) = 2.8944(0.9(17)+10)^{-0.7} =0.3015...\approx 0.30$

Contextually, this means the percetange is increasing by about 0.30% per year.

Part C)

For this question, we will simply use the original function since it outputs the percentage of the American population 55 and older. Thus, evaluate f(t) when t = 17:

$\displaystyle f(17) = 10.72(0.9(17)+10)^{0.3}=28.2573...\approx 28.26$