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4 years ago

How many hairs does a dog have

Mathematics

1 answer:

Morgarella [4.7K]4 years ago

8 0

Answer:

$15,000~hairs$

Step-by-step explanation:

$A~dog~has~about~15,000~hairs~on~each~little~square~inch~of~skin.$

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ANSWER ASAP FOR BRAINLIESTT

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3 years ago

Find the volume of each sphere, both in terms of pi and to the nearest tenth. use 3.14 for pi

zaharov [31]

V=4/3*3.14*81
V=339,12 cubic ft

6 0

4 years ago

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Evaluate: (6 + 2)^3-12 5

Archy [21]

Answer:

=100 i am 100% sure lol

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4 years ago

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our goal is to run a mean of 45 minutes per day for a week. For the first six days, you run 38, 40, 40, 42, 43, and 50 minutes.

oksano4ka [1.4K]

Answer:

62 minutes.

Step-by-step explanation:

To find the mean, you need to add together all of your values (the minutes) and divide them by the number of values (how many sets of minutes given).

So, choosing to add 62 minutes:

38 + 40 + 40 + 42 + 43 + 50 + 62 = 315

There are 7 sets of minutes in total, so we then divide 315 by 7.

315/7 = 45 minutes

Therefore, running 62 minutes on the seventh day would cause you to run a mean of 45 minutes per day for the week.

6 0

3 years ago

Integrate dx/3sinx+4cosx

german

$\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}$

A standard approach would be the tangent half-angle substitution:

$t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx$

Then

$\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}$

$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}$

from which we get

$\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt$

So the integral becomes

$\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}$

Rewrite the denominator as

$2t^2-3t-2=(2t+1)(t-2)$

and expand the integrand into its partial fractions:

$\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)$

We have

$\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C$

$=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C$

$=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C$

6 0

3 years ago