All categories

3 years ago

What happens to the volume of a sphere when its diameter is doubled?

2 answers:

6 0

Consider this solution:
1) formula of volume is $V= \frac{4}{3} \pi r^3, where \ r - radius$
2) if new_r=2r, then the formula is $new_V= \frac{4}{3}*(2r)^3= \frac{4}{3} \pi *8r^3=8*( \frac{4}{3} \pi r^3)$
3) new_V/V=8:1
The volume increases by 8 times.

Taya2010 [7]3 years ago

3 0

Answer:

Its volume is 8 times greater.

Step-by-step explanation:

I haveTTM for the users

You might be interested in

Find the sum of 2 and 3/2

poizon [28]

Answer:<u>3</u>

Step-by-step explanation:2+3/2=

3/2=1

2+1=3 Or 3.5

8 0

3 years ago

Read 2 more answers

12(x-3)+18= what is the awnser

Mila [183]

Answer:

Step-by-step explanation:

The answer would be 32:);)

7 0

3 years ago

Fico is a credit score. true false user: collin has a credit card bill of $3,000. he makes only the minimum payment and is alway

Alenkasestr [34]

Fico is a credit score - true
Collin is not using credit responsibly.

I don't think my answer is correct but I hope it helped you :)

6 0

3 years ago

Read 2 more answers

First answer gets brainliest. And one hundo points

enyata [817]

Answer:The one you already have marked,B.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Find the minimum and maximum values of the function subject to the given constraint. (if an answer does not exist, enter dne.) f

nata0808 [166]

Via Lagrange multipliers:

$L(x,y,\lambd)=x^2y+x+y+\lambda(xy-5)$
$L_x=2xy+1+\lambda y=0$
$L_y=x^2+1+\lambda x=0$
$L_\lambda=xy-5=0$

$\underbrace{10}_{2xy}+1+\lambda y=0\implies \lambda=-\dfrac{11}y$
$xy=5\implies y=\dfrac5x\implies\lambda=-\dfrac{11}5x$

$\impliesx^2+1+\left(-\dfrac{11}5x\right)x=0\implies x^2=\dfrac56\implies x=\pm\sqrt{\dfrac56}$
$xy=5\implies y=\pm\sqrt{30}$

At these points, we get local minima of $f\left(\pm\sqrt{\dfrac56},\pm\sqrt{30}\right)=\pm2\sqrt{30}$ .

- - -

Another way to do this is to make $f(x,y)$ a function independent of $y$ , which is made possible by the constraint.

$xy=5\implies y=\dfrac5x$
$\implies f(x,y)=f\left(x,\dfrac5x\right)=F(x)=6x+\dfrac5x$
$\implies F'(x)=6-\dfrac5{x^2}=0\implies x=\pm\sqrt{\dfrac56}$

and so on.

8 0

3 years ago