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3 years ago

10

Someone answer this correctly please!!

1 answer:

Sidana [21]3 years ago

8 0

Answer:

8 < $\sqrt{71}$ < 9

Step-by-step explanation:

You need to find the perfect squares above and below 71

8*8= 64

9*9 =81

8 < $\sqrt{71}$ < 9

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Solve -(-7)^2+6(2)^2

Lilit [14]

-(-7)^2+6(2)^2
-(49)+6(4)
-49+24
-25

6 0

3 years ago

HELP NEED ANSWERS IN AT LEAST 5 MINUTES Do prokaryotes organisms contant a nucleus? O Yes O No

KATRIN_1 [288]

Step-by-step explanation:

No prokaryotes organisms doesnot contant a nucleus

Mark as brainliest...

5 0

3 years ago

Read 2 more answers

The base of a solid is the region in the first quadrant between the graph of y=x2 and the x -axis for 0≤x≤1 . For the solid, eac

nata0808 [166]

Answer:

The volume of the solid is π/40 cubic units.

Step-by-step explanation:

Please refer to the graph below.

Recall that the area of a semi-circle is given by:

$\displaystyle A=\frac{1}{2}\pi r^2$

The volume of the solid will be the integral from <em>x</em> = 0 to <em>x</em> = 1 of area A. Since the diameter is given by <em>y</em>, then the radius is <em>y/2</em>. Hence, the volume of the solid is:

$\displaystyle V=\int_0^1\frac{1}{2}\pi \left(\frac{y}{2}\right)^2\, dx$

Substitute:

$\displaystyle V=\frac{1}{2}\pi\int_0^1\left(\frac{x^2}{2}\right)^2\, dx$

Simplify:

$\displaystyle V=\frac{1}{2}\pi \int_0^1\frac{x^4}{4}\, dx$

Integrate:

$\displaystyle V=\frac{1}{2}\pi \left[\frac{x^5}{20}\Big|_0^1\right]$

Evaluate:

$\displaystyle V=\frac{\pi}{40}\left((1)^5-\left(0\right)^5\right)=\frac{\pi}{40}\text{ units}^3$

The volume of the solid is π/40 cubic units.

4 0

3 years ago

A right angle has an area of 18 square inches.if the triangle has two sides that are the same length, what are the lengths of th

NeX [460]

I hope this helps you

this triangle is isosceles right triangle

base=height

Area=base×height /2

18=b×b/2

b^2=36

b=6

8 0

3 years ago

Y=x.arctan(x)^1/2. find dy/dx. pls show steps

Whitepunk [10]

$y=x(\arctan x)^{1/2}$

Use the product rule first:

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dx}{\mathrm dx}(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}$

$\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}$

Use the chain rule to compute the derivative of $(\arctan x)^{1/2}$ . Let $z=(\arctan x)^{1/2}$ and take $u=\arctan x$ , so that by the chain rule

$\dfrac{\mathrm dz}{\mathrm dx}=\dfrac{\mathrm dz}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm dz}{\mathrm du}=\dfrac{\mathrm du^{1/2}}{\mathrm du}=\dfrac12u^{-1/2}$

$\dfrac{\mathrm du}{\mathrm dx}=\dfrac{\mathrm d\arctan x}{\mathrm dx}=\dfrac1{1+x^2}$

$\implies\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}(1+x^2)}$

So we have

$\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+\dfrac x{2(\arctan x)^{1/2}(1+x^2)}$

You can rewrite this a bit by factoring $(\arctan x)^{-1/2}$ , just to make it look neater:

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}}\left(2\arctan x+\dfrac x{1+x^2}\right)$

3 0

3 years ago