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Nikolay [14]
2 years ago
6

How do I solve this.

Mathematics
1 answer:
blondinia [14]2 years ago
3 0

You are asked to find the error. The error is that 4,100,000 is the answer to 4.1×10^6, not 4.1×10^-6. The correct answer to 4.1×10^-6 is 1/4,100,000.

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1/3 of a number is 12 less than the number itself
natulia [17]
Let "x" be a substitute for "a number"

1/3(x)=x-12    Multiply by 3
x=3x-36         Get n onto the same side of the equation
2x=36            Divide by 2
x=18

That's your answer.
5 0
3 years ago
A bag contains 10 white golf balls and 6 striped golf balls. A golfer wants to add 112 golf balls to the bag. He wants the ratio
dem82 [27]
Ratio 10:6
so 10+6 = 16

112/16 =7

white balls: 10 * 7 = 70
striped balls: 6 * 7 = 42

Answer: he needs to add 70 white balls and 42 striped balls to keep the same ratio 10:6
8 0
3 years ago
The base of S is the region enclosed by the parabola y = 8 − 8x2 and the x-axis. Cross-sections perpendicular to the x-axis are
Elena-2011 [213]

The base of a solid is the region in the first quadrant bounded by the y-axis, the x-axis, the graph of y=ex, and the vertical line x=1. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume of the solid?

Step-by-step explanation:

Answer: Sometimes I dont want to be happy.

8 0
1 year 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
For what values of x is the quantity x−9 negative?
Contact [7]
X - 9 < 0
add 9 to both sides:-
x - 9 + 9   < 9
x < 9

x - 9 is negative for all values of x less that 9.
6 0
3 years ago
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