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3 years ago

I’m really stupid. I need help

Mathematics

2 answers:

Gnom [1K]3 years ago

7 0

Answer:

7. 30 8. 1.89

Step-by-step explanation:

30/1=30

9.45/5=1.89

patriot [66]3 years ago

3 0

Answer:

1 box per 30 pounds

1 notebook per 1.89 dollars

Step-by-step explanation:

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What is the smallest power of ten that would exceed 987654321098765432

Romashka [77]

10 to the power of 18

3 0

3 years ago

an indy car driver has to be a certain height to fit into the race car. The following inequality 175<=3x-17<=187, where x

enot [183]

Given inequality : 175 ≤ 3x-17 ≤ 187, where x represents the height of the driver in inches.

Let us solve the inequality for x.

We have 17 is being subtracted in the middle.

Reverse operation of subtraction is addition. So, adding 17 on both sides and also in the middle, we get

175+17 ≤ 3x-17+17 ≤ 187+17

192 ≤ 3x ≤ 204.

Dividing by 3.

192/3 ≤ 3x/3 ≤ 204/3.

64 ≤ x ≤ 68.

Therefore, the height of the driver should be from 64 to 68 inches to fit into the race car.

8 0

3 years ago

Can someone help me on this math problem ?

faltersainse [42]

Answer:

4/3

Step-by-step explanation:

Given two points, we can find the slope by using

m = (y2-y1)/(x2-x1)

= (20 - -16)/(7 - -20)

= (20+16)/(7+20)

= 36/27

=4/3

8 0

3 years ago

Gianna is converting a 12-foot-by-15-foot room in her

givi [52]

Answer:

is there a picture of the question

5 0

2 years ago

A right circular cone is undergoing a transformation in such a way that the radius of the cone is increasing at a rate of 1/2 in

Ivenika [448]

Answer:

The volume is decreasing at the rate of 1.396 cubic inches per minute

Step-by-step explanation:

Given

Shape: Cone

$\frac{dr}{dt} =\frac{1}{2}$ --- rate of the radius

$\frac{dh}{dt} =-\frac{1}{3}$ --- rate of the height

$r = 2$

$h = \frac{1}{3}$

Required

Determine the rate of change of the cone volume

The volume of a cone is:

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

Differentiate with respect to time (t)

$\frac{dV}{dt} = \frac{\pi}{3}(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})$

Substitute values for the known variables

$\frac{dV}{dt} = \frac{\pi}{3}(2*2*\frac{1}{3}* \frac{1}{2} - 2^2 *\frac{1}{3})$

$\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}* \frac{1}{2} - \frac{4}{3})$

$\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}(\frac{1}{2} - 1))$

$\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}*- 1)$

$\frac{dV}{dt} = -\frac{\pi}{3}*\frac{4}{3}$

$\frac{dV}{dt} = -\frac{22}{7*3}*\frac{4}{3}$

$\frac{dV}{dt} = -\frac{22}{21}*\frac{4}{3}$

$\frac{dV}{dt} = -\frac{88}{63}$

$\frac{dV}{dt} =-1.396in^3/min$

The volume is decreasing at the rate of 1.396 cubic inches per minute

3 0

3 years ago