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exis [7]
3 years ago
11

A rectangular prism has a base that is 4 m by 6 m and a height of 10 m. If all dimensions are doubled, what happens to the volum

e? Explain the steps that you take to arrive at your answer.
Mathematics
2 answers:
Anastasy [175]3 years ago
6 0

Answer:

Start by calculating the volume of the original prism. The area of the base, B, is 4(6) = 24 m2. Multiply that by the height to get the volume: 24(10) = 240 m3. Now, if you double all of the dimensions, you would have a rectangular prism with a base that is 8 m by 12 m and a height of 20 m. The area of the base, B, is 8(12) = 96 m2. Multiply that by the height to get the volume: 96(20) = 1,920 m3. To compare, look at the ratio of the original prism to the new prism: 96/1,920 = 1/8. The volume increases by 8 times the original volume.

Step-by-step explanation:

Sample response

nirvana33 [79]3 years ago
3 0
When the dimensions of a rectangular prism will be doubled, its volume won't only be doubled but it will be 8 times as much.

The original volume of the rectangular prism is LxWxH.

V= LxWxH
V= (6m) (4m) (10m)
V= 240m³

If we double the dimensions, the volume would be

V= 2L x 2W x 2H
V= 2 x 6m x 2 x 4m x 2 x 10m
V= 2 x 2 x 2 x 6m x 4m x 10m
V= 2³ (6m x 4m x 10m)
V= 2³ x 240m³
V = 8 x 240m³
V= 1920 m³

The new volume now is 1,920 m³, which makes the rectangular prism larger than the original.
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3 years ago
What time is it if half of what has already passed remains until the end of today?
lukranit [14]

Answer:

<h2>4 p.m.</h2>

Step-by-step explanation:

We know that a day contains 24 hours in total.

The time already passed will be represented by x.

The remaining time would be \frac{x}{2}, becasye half of what has already passed remains until the end of the day.

Basically, the sum of these two expression gives 24 hours in total.

x + \frac{x}{2} =24\\\frac{3x}{2}=24\\ x=16

Therefore, the actual time is 4 p.m.

7 0
3 years ago
Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |
dsp73

Looks like f(x,y)=x^2+y^2-x^2y+7.

f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1

f_y=2y-x^2=0\implies2y=x^2

  • If x=0, then y=0 - critical point at (0, 0).
  • If y=1, then x=\pm\sqrt2 - two critical points at (-\sqrt2,1) and (\sqrt2,1)

The latter two critical points occur outside of D since |\pm\sqrt2|>1 so we ignore those points.

The Hessian matrix for this function is

H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}

The value of its determinant at (0, 0) is \det H(0,0)=4>0, which means a minimum occurs at the point, and we have f(0,0)=7.

Now consider each boundary:

  • If x=1, then

f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4

which has 3 extreme values over the interval -1\le y\le1 of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

  • If x=-1, then

f(-1,y)=8-y+y^2

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

  • If y=1, then

f(x,1)=8

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

  • If y=-1, then

f(x,-1)=2x^2+8

which has 3 extreme values, but the previous cases already include them.

Hence f(x,y) has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0
3 years ago
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