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3 years ago

PLEASEEEE HELP ME WITH THIS ALGEBRA 1

Mathematics

1 answer:

AleksandrR [38]3 years ago

5 0

Answer:

Step-by-step explanation:

total cost = tickets( c) +50

1400 = 4 (c) + 50

1350 = 4 (c)

337.50 = c

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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

2 years ago

PLZ HELP ASAP AND SHOW UR WORK!!!! THE ≤. WITHOUT THE LINE IN THE BOTTOM

Marina86 [1]

Answer:

Step-by-step explanation:

$\frac{2}{3}x +6$

subtract 6 from both sides

$\frac{2}{3}x$

divide by 2/3(or multiply by 3/2)

$x$

and thats your answer

3 0

3 years ago

What's the error? Jerry says that a cube with edges that measure 10 cm has a volume that is twice as much as a cube with sides t

disa [49]

The quick way to dispute something like this is to simply do the calculation and then create a ratio.

Cube One (Large Cube)
The formula for a cube is V = e^3
e = the measurement of an edge. In this case.
e = 10 cm
V = e^3
V = 10^3 = 10*10*10
V = 1000 cm^3

Cube 2 (Small Cube)
V = e^3
e = 5 cm
V = 5*5*5
V = 125 cm^3

Ratio
Large Cube / Small Cube = 1000 / 125 = 8/1.

The difference in size is 8 to 1 not 2 to 1.

Explanation
He's right if he sticks to one side. The ratio of one side of the large cube to the small one is 2 to 1. But once you put that into the formula for volume, three sides are multiplied together and that 2 shows up everytime you multiply the sides together.

4 0

3 years ago

A cable company charges a flat fee of $50 per month plus $8 per month for each movie channel, m, and $4 per month for each sport

PSYCHO15rus [73]

T = 12( 50 + 8m + 4s)

T = 600 + 96m + 48s

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3 years ago

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How to do q44<br>thx!!!!

aleksklad [387]

1. B
Adding a constant offset to the data does not change its variance. Scaling it, or applying a non-linear function will change the variance.

2. A
For a given number on the 3rd die, the first two dice can match it 11 ways, or have two numbers that match in 5 additional ways. Thus probability of 2 or more numbers being the same is 6*16/216 = 4/9.

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3 years ago