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olya-2409 [2.1K]
3 years ago
6

Help. I need both. Please

Mathematics
1 answer:
Darya [45]3 years ago
6 0
Anything to the power of 0 is 1

So for the top, 2^0 = 1
For the bottom, 10^0 = 1


It is not 0, because anything that ends with 0/0 is <em>undetermined</em>, and will not result in an answer

hope this helps
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x1= -5 x2=0.5 y= -5

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Step-by-step explanation:

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a bathtub is in the shape of a rectangular prism. its dimensions are 5 feet by 3 feet by 18 inches. the bathtub is three-fourths
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The bathtub has dimensions 5 ft by 3 ft by 18 inches.
Note that 18 inches = 18/12 = 1.5 ft.

The volume of the bathtub is
V = 5*3*1.5 = 22.5 ft³

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3 years ago
If f(x) = |x| + 9 and g(x) = –6, which describes the range of (f + g)(x)?
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A cylinder has a volume of 245 cubic units and a height of 5 units. The diameter of the cylinder is
Vikki [24]

Answer:

The diameter is twice that, or approx. 7.90 units.

Step-by-step explanation:

the equation for the volume of a cylinder of radius r and height h is

V = πr²h.  Here we need to calculate the diameter after having found the radius.  Solving V = πr²h for r², we get:

           V

r² = -----------

         πh

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         145 units³

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       3.14159(5 units)

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3 0
3 years ago
Read 2 more answers
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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