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2 years ago

11) the difference between the product of 4 and a number and the square of the number

Mathematics

1 answer:

marta [7]2 years ago

6 0

Answer:

$= 4x-x^2\\\\$

Step-by-step explanation:

Let the number be x;

The product of 4 and the number will be;

$= 4 * x\\\\= 4x$

The the square of the number is given as;

$= x^2\\$

Taking the difference of both expressions above;

$= 4x-x^2\\\\$

Factor out the common term:

$= x(4-x)\\\\$

Hence the difference between the product of 4 and a number and the square of the number is $= 4x-x^2\\\\$

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A spherical balloon currently has a radius of 19cm. If the radius is still growing at a rate of 5cm or minute, at what rate is a

miv72 [106K]

Answer:

22670.8 cm³/min

Step-by-step explanation:

Given:

Radius of the balloon at a certain time (r) = 19 cm

Rate of growth of radius is, $\frac{dr}{dt}=5\ cm/min$

The rate at which the air is pumped in the balloon can be calculated by finding the rate of increase in the volume of the balloon.

So, first we find the volume of the sphere in terms of 'r'. As the balloon is spherical in shape, the volume of the balloon is equal to the volume of a sphere. Therefore,

Volume of balloon is given as:

$V=\frac{4}{3}\pi r^3$

Now, rate of increase of volume is obtained by differentiating both sides of the equation with respect to time 't'.

Differentiating both sides with respect to time 't', we get:

$\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4\pi}{3}(3r^2)(\frac{dr}{dt})\\\\\frac{dV}{dt}=4\pi r^2(\frac{dr}{dt})$

Now, plug in 19 cm for 'r', 5 cm per minute for $\frac{dr}{dt}$ and solve for $\frac{dV}{dt}$ . This gives,

$\frac{dV}{dt}=4\pi (19 cm)^2(5\ cm/min)\\\\\frac{dV}{dt}=4\times 3.14\times 361\times 5\ cm^3/min\\\\\frac{dV}{dt}=22670.8\ cm^3/min$

Therefore, the rate at which the air is being pumped into the balloon is 22670.8 cm³/min.

4 0

2 years ago

There are 100 people in a sport centre.

Arlecino [84]

We need to determinate who doesn't use anything, so first we must know the real number of who are practising: a person can do more things.

For doing this, is better use a three-circles graphic, but we try to do without it.

16 pool + gym + track, and this number is sure

38 gym + pool, but this number includes the people of 1st point. We must know who do only gym + pool: 38 - 16 = 22

31 pool + track, this number also includes people of 1st point, so: only pool + track 31 - 16 = 15

33 gym + track, this number also includes people of 1st point, so: only gym + track 33 - 16 = 17

now:

67 gym, we must know who use only gym: 67 - 16 (gym + pool + track) - 22 (gym + pool) - 17 (gym + track) = 12

62 pool, we must know who use only pool: 62 - 16 (gym + pool + track) - 22 (gym + pool) - 15 (pool + track) = 9

56 track, we must know who use only track: 56 - 16 (gym + pool + track) - 15 (pool + track) - 17 (gym + track) = 8

now we must now who do some facility: let's sum all

16 + 22 + 15 + 17 + 12 + 9 + 8 = 99

99/100 practise something

So only 1/100 doesn't use nothing

The probability is 1%

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2 years ago

I can't even read the two way table please help

Ne4ueva [31]

It's C. Number of students who scored "proficient" on chapter 7 is 16, and of those, only 3 scored "needs improvement" on 6. So the 3/16 represents those who were proficient at 7, but need improvement on 6

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2 years ago

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labwork [276]

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2 years ago

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