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2 years ago

Help this may be the last time

Mathematics

1 answer:

Anit [1.1K]2 years ago

3 0

I think the first is 107 and the second is 2. i’m not sure about the last two blanks but if they’re asking if it’s equivalent or not i think it is

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Help plz I’m not good at this

Natalka [10]

Answer:

The correct answer is A. 1/8

Step-by-step explanation:

Let's solve (1/3 * 9/16) / 1 1/2 = x

(1/3 * 9/16) / 1 1/2 = x

9/48 / 1 1/2 = 9/48 * 2/3 = 18/144

18/144, we can simplify dividing by 18 at the numerator and the denominator, this way:

x = 18 ÷ 18/ 144 * 18 = 1/8

The correct answer is A. 1/8

4 0

2 years ago

4 Solve these inequalities. Show the solutions on the number lines.

Roman55 [17]

Answer:

a) v>4.5

b) w<6.5

a) 0.3 recurring

b) 5/11

6 0

1 year ago

The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi

inessss [21]

Answer:

The height of cone is decreasing at a rate of 0.085131 inch per second.

Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

$\displaystyle\frac{dr}{dt} = -7\text{ inch per second}$

The volume is decreasing at a constant rate.

$\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}$

Instant radius = 99 inch

Instant Volume = 525 cubic inches

We have to find the rate of change of height with respect to time.

Volume of cone =

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

Instant volume =

$525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}$

Differentiating with respect to t,

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

Putting all the values, we get,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131$