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FrozenT [24]
3 years ago
13

(-3y2 + 2y + 9) (7y - 2)​

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
8 0

Answer:

simplified is -22y^3+20y^2+59y-18

Step-by-step explanation:

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vladimir1956 [14]
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Instructions: follow the bedmas rule of solving.
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Shiraz want to create an equally tricky arithmetic sequence she wants the 5th term of the sequence to equal 11 and 50th Cham to
docker41 [41]

i dont have an idea :)

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3 years ago
4. a) A ping pong ball has a 75% rebound ratio. When you drop it from a height of k feet, it bounces and bounces endlessly. If t
Klio2033 [76]

First part of question:

Find the general term that represents the situation in terms of k.

The general term for geometric series is:

a_{n}=a_{1}r^{n-1}

a_{1} = the first term of the series

r = the geometric ratio

a_{1} would represent the height at which the ball is first dropped. Therefore:

a_{1} = k

We also know that the ball has a rebound ratio of 75%, meaning that the ball only bounces 75% of its original height every time it bounces. This appears to be our geometric ratio. Therefore:

r=\frac{3}{4}

Our general term would be:

a_{n}=a_{1}r^{n-1}

a_{n}=k(\frac{3}{4}) ^{n-1}

Second part of question:

If the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces.

k represents the initial height:

k = 235\ ft

n represents the number of times the ball bounces:

n = 6

Plugging this back into our general term of the geometric series:

a_{n}=k(\frac{3}{4}) ^{n-1}

a_{n}=235(\frac{3}{4}) ^{6-1}

a_{n}=235(\frac{3}{4}) ^{5}

a_{n}=55.8\ ft

a_{n} represents the highest height of the ball after 6 bounces.

Third part of question:

If the ball dropped from a height of 235ft, find the total distance traveled by the ball when it strikes the ground for the 12th time. ​

This would be easier to solve if we have a general term for the <em>sum </em>of a geometric series, which is:

S_{n}=\frac{a_{1}(1-r^{n})}{1-r}

We already know these variables:

a_{1}= k = 235\ ft

r=\frac{3}{4}

n = 12

Therefore:

S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{1-\frac{3}{4} }

S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{\frac{1}{4} }

S_{n}=(4)(235)(1-\frac{3}{4} ^{12})

S_{n}=910.22\ ft

8 0
3 years ago
The volume of a right circular cone with radius r and height h is V = pir^2h/3.
Scorpion4ik [409]

The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

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

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

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9  and dr = 6.8 - 5.9 = 0.9

     h = 4  and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b).  The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47  and dr = 6.45 - 6.47 = -0.02

     h = 10  and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

8 0
3 years ago
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