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kherson [118]
2 years ago
11

Hurry it's a math problem

Mathematics
2 answers:
svet-max [94.6K]2 years ago
6 0
3x3 is 9 4x4 is 16 (right triangle) so 90-25 = x (65)
eimsori [14]2 years ago
4 0
Do 1. It’s a hypotenuse so do 1 and find x
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What is a equivalent rational number to 1.28 repeating?
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<span>the line over the 28 means the 28 repeats forever. 1.282828.... and so on let x be the rational number 1.28... we can use this trick: 100*1.282828....= 128.282828... (the decimal 28 part repeats) 100x = 128.28... next: 100x - x = 128.282828... - 1.282828... the .282828... part will be subtracted away 99x = 127 divide both sides by 99 to get x= 127/99</span>
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2 years ago
Mrs shirman has a bulletin that is 6 feet and 5 feet wide. She has 32 feet of border to go around the edges of the board. Does s
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8 0
2 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
2 years ago
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