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2 years ago

Find the square root of each by division method:

Mathematics

2 answers:

Dima020 [189]2 years ago

8 0

Refer to the attachment

Dafna1 [17]2 years ago

6 0

Answer:

$\sqrt[2]{225} =15$ , $\sqrt[2]{961} =31$ , $\sqrt[2]{3969} =69$ , $\sqrt[2]{10404} =102$

Step-by-step explanation:

$\sqrt[2]{225} =\sqrt{15*15}=15$

$\sqrt[2]{961}=\sqrt{31*31} =31$

$\sqrt[2]{3969} =\sqrt{69*69} =69$

$\sqrt[2]{10404} =\sqrt{102*102}=102$

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In a 10 m2 ecosystem, there are 20 raccoons. The population density of raccoons is _____.

katrin2010 [14]

To find the density of racon all you have to do is to divide number of racoons with surface area because as name says you need to get x number of racoons per m^2

its own name tells you what you need to divide with what.

answer is:
20/10 = 2 raccoons per m^2

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2 years ago

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Marcus is building a fence around his rectangular backyard. The length of his backyard is 75 3/4 ft, and the width is 81 1/4 ft.

Allisa [31]

Answer:

Step-by-step explanation:

Answer:

4,923.75ft²

Step-by-step explanation:

The area of the backyard = Length * width \

Area of the backyard = 75 3/4 * 81 1/4

Area of the backyard = 303/4 * 325/4

Area of the backyard = 6,154.6875ft²

If 3/4 of the fence is finished, the area of the finished part is expressed as;

Finished part = 1/5 * 6,154.6875

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Part left unfinished = 6,154.6875-1,230.9375

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4 0

3 years ago

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According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

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

25 points ???? ASAP PLEASE HELP

ArbitrLikvidat [17]

Answer:

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Step-by-step explanation:

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2 years ago

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Maru [420]

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2 years ago

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