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3 years ago

7

Please help me with this. i don't understand it.

2 answers:

serg [7]3 years ago

5 0

Answer:

Step-by-step explanation:

1) Rectangular prism: l = 7.5 cm ; w = 12 cm ; h = 15.5cm

Volume of rectangular prism = l * w * h

Cubes: edge = 0.5cm

Volume of cube = edge *edge * edge

Number of cubes = Volume of rectangular prism ÷ Volume of a cube

$=\frac{7.5 * 12 * 15.5}{0.5*0.5*0.5}\\\\ = \frac{7.5 * 12 * 15.5*1000}{0.5 *0.5 *0.5 *1000}\\\\= \frac{75 * 12 *155 * 10}{5 * 5 * 5}\\\\= 3 * 12 * 31 * 10\\\\= 11160$

11160 cubes will fit in the prism.

2) Volume of prism = 7.5 * 12 * 15.5

= 1395 cm³

Vika [28.1K]3 years ago

4 0

Answer:

A. 11160 cubes B. 1395cm^3

Step-by-step explanation:

Volume= length x width x height

Its easiest to solve B. first and use it to solve A.

B. 7.5 x 12 x 15.5 = 1395cm^3 (volume of prism)

A. .5 x .5 x .5 = 1/8cm^3 (which is the volume of the cube)

1395 ÷ 1/8 = 11160

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Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

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$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

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b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

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$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

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