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charle [14.2K]
3 years ago
13

the area of a children’s square playground is 50m squared, what is the exact length of the playground

Mathematics
2 answers:
butalik [34]3 years ago
6 0

Answer:

\displaystyle 5\sqrt{2}\:m.

Step-by-step explanation:

Sinse squares are equal in length all around, this formula applies:

\displaystyle s^2 = A → \sqrt{50} = \sqrt{s^2} \\ \\ 5\sqrt{2} = s

Therefore, the length of the playground is \displaystyle 5\sqrt{2} metres.

* Just in case you could not figure out how to find the square root of fifty, here is how using perfect squares:

\displaystyle \sqrt{50} → \sqrt{2 \times 25} \\ \\ 5\sqrt{2}

Sinse we are dealing with measurements, we only want the NON-NEGATIVE root.

I am joyous to assist you at any time.

tensa zangetsu [6.8K]3 years ago
4 0

Area=50m^2

Let side be a

\\ \rm\Rrightarrow Area=a^2

\\ \rm\Rrightarrow a^2=50

\\ \rm\Rrightarrow a=\sqrt{50}

\\ \rm\Rrightarrow a=7.1m

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Ms. Fontanges wants to put a line of bricks along the back edge of her yard. She needs 35 feet of bricks. The bricks she wants t
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5 0
2 years ago
According to data from a medical​ association, the rate of change in the number of hospital outpatient​ visits, in​ millions, in
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Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

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3 years ago
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