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marishachu [46]

3 years ago

8

Find the volume of a right circular cone that has a height of 4.1 cm and a base with a diameter of 14.8 cm. Round your answer to

the nearest tenth of a cubic centimeter.

1 answer:

podryga [215]3 years ago

6 0

Answer:

$V=235.1\ cm^3$

Step-by-step explanation:

<u>Volume of a Cone</u>

The volume of a right circular cone with height h and a base radius r is:

$\displaystyle V=\frac{1}{3}\pi r^2 h$

The height of the given cone is h=4.1 cm and its diameter is d=14.8 cm. Since the radius is half the diameter, r=7.4 cm.

Calculate the volume:

$\displaystyle V=\frac{1}{3}\pi\cdot 7.4^2\cdot 4.1$

$\displaystyle V=\frac{1}{3}\pi\cdot 54.76\cdot 4.1$

$\displaystyle V=\frac{1}{3}\pi\cdot 224.516$

$\boxed{V=235.1\ cm^3}$

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Vector u has a magnitude of 4 with a direction of 50°. Vector v has a magnitude of 6 with

Luda [366]

Answer: <8.01, 5.60>.

Step-by-step explanation:

Vector:

$v= (r\cos\theta,r\sin\theta)$

, where r = magnitude and $\theta$ = angle .

As per given,

$u=(4\cos 50^{\circ}, 4\sin 50^{\circ})\\\\\approx(4(0.64),4(0.77))\\\\=(2.56, 3.08)$

$v=(6\cos 25^{\circ}, 6\sin 25^{\circ})\\\\\approx(6(0.91),6(0.42))\\\\=(5.46, 2.52)$

Consider

$u+v=(2.56,3.08)+(5.46,2.52)\\\\=(2.56+5.46,3.08+2.52)\\\\=(8.02,\ 5.60)$

Hence, the correct option is <8.01, 5.60>.

7 0

3 years ago

expeople1 [14]

ANSWER 1

Note that,

$f(u)=tan^{-1}(u)$

is the same as

$f(u)=arctan(u)$

We apply the product rule.

$f(x)=x^2tan^{-1}(x)$

So we keep the second function and differentiate the first,plus we keep the first function and differentiate the second.

$f'(x)=(x^2)'tan^{-1}(x)+x^2(tan^{-1}(x))'$

Recall that,

If

$f(u)=tan^{-1}(u)$

Then,

$f'(u)=\frac{1}{1+u^2}} \times u'$

This implies that,

$f'(x)=2xtan^{-1}(x)+\frac{x^2}{x^2+1}$

ANSWER 2

We apply the product rule and the chain rules of differentiation here.

$f(x)=xsin^{-1}(1-x^2)$

$f'(x)=x'sin^{-1}(1-x^2)+x(sin^{-1}(1-x^2))'$

Recall that,

If

$f(u)=sin^{-1}(u)$

Then,

$f'(u)=\frac{1}{\sqrt{1-u^2}} \times u'$

This implies that,

$f'(x)=sin^{-1}(1-x^2)+x \times \frac{1}{\sqrt{1-(1-x^2)^2}}\times (-2x)$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-(1-2x^2+x^4)}}$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-1+2x^2-x^4}}$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{2x^2-x^4}}$

5 0

4 years ago

6 + ( 2 − 3 ( 4 ) ) + 1 /( 5 ) ( 2 ) − 4

tankabanditka [31]

6 + (2 - 3 (4) )+1 / (5) (2) - 4 = -1 /2

<u>Step-by-step explanation:</u>

6 + (2 - 3 (4) )+1 / (5) (2) - 4

Using the BODMAS rule, we can simplify the expression in the following way, like, we have to do the operation inside the brackets first, then division, multiplication, addition and then subtraction.

= 6 + (2 - 3 (4) )+1 / (5) (2) - 4 [operation inside the brackets]

= 6 + (2 - 12) )+1 / (5) (2) - 4

= 6 + (-10) )+1 / (10) - 4

= 6 + 1 -10 / (10) - 4

= 7 - 10/ 6

= -3 / 6

= -1/2

4 0

3 years ago

nikdorinn [45]

A is one correct answer

8 0

3 years ago

There are 52 weeks in a year. If Kendra runs 12 miles every week, how many miles will she run in a year?

Bumek [7]

Okay so there are 52 weeks in a year and she runs 12 miles every week so multiply 52 times 12 which equals 624

6 0

4 years ago

Read 2 more answers