- 7.5 divided by

The width of a swimming pool is one third of its length.The width of the pool is 15 feet.What is the length of the pool?Write an

zubka84 [21]

The answer is 45

Here is why: 15*3=45

5 0

3 years ago

Write the equation of the line in slope intercept form that goes through the point (2,3) and has a slope of -1/2.

elena55 [62]

The answer is y =-1/2x + 3

4 0

4 years ago

What is 19.2 * 10^8 minutes in hours? Written in scientific notation form?

Lena [83]

Answer should be 1.85 x 10^18

7 0

3 years ago

I don't really understand the question so can someone help me answer the question please?

Ksivusya [100]

Answer:

1/ 3

Step-by-step explanation:

If each of the cards is turned over, the probability of picking up a card of one type P(E) becomes equal to:

=> P(E) = number of cards of the required type/ total number of cards

● Total number of spades( ♤ ) = 3

{the queen, one ace and the nine are all spades}

● Total number of cards = 6

Probability of drawing a spade= 3/ 6

= 1/ 2

● Total number of "7" = 1

● Total number of cards = 6

Probability of drawing a 7

= 1/ 6

Now, what's asked is the difference in the probabilities of drawing a spade and a seven.

= 1/ 2 - 1/ 6

= 3/ 6 - 1/ 6

= 2/ 6

= 1/ 3

Hence, 1/ 3 of a greater chance of drawing a spade over a 7.

5 0

3 years ago

Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6

Anton [14]

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

V= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder, $x^{2}+y^{2} =1$ and the ellipsoid, $4x^{2}+ 4y^{2} + z^{2} =64$

In polar coordinates, we know that

$x^{2}+y^{2} =r^{2}$

So, the ellipsoid gives

$4{(x^{2}+ y^{2)} + z^{2} =64$

4( $r^{2}$ ) + $z^{2}$ = 64

$z^{2}$ = 64- 4( $r^{2}$ )

z=± $\sqrt{64-4r^{2} }$

So, the volume of the solid is given by:

V= $\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta$

= $2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta$

To solve the integral take, $64-4r^{2}$ = t

dt= -8rdr

rdr = $\frac{-1}{8} dt$

So, the integral $\int\ r\sqrt{64-4r^{2} } rdr$ become

= $\int\ \sqrt{t } \frac{-1}{8} dt$

= $\frac{-1}{12} t^{3/2}$

= $\frac{-1}{12} (64-4r^{2}) ^{3/2}$

so on applying the limit, the volume becomes

V= $2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta$

= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta$

V = $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Since, further the integral isn't having any term of $\theta$ .

we will end here.

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

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3 0

2 years ago

- 7.5 divided by - 2.5