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3 years ago

a package shaped like a cube has an edge that is 28 cm long. How much space is available to pack inside your box?

Mathematics

1 answer:

Tema [17]3 years ago

7 0

Answer: 21,952 cm³

Explanation:

Volume = 28 x 28 x 28 = 29 152 cm³

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A - g(x) = |x-5|+5 <br> B - g(x) = |x-5| <br> C - g(x) = |x|+5 <br> D - g(x) = |x+5|+5

n200080 [17]

Answer:

$g(x) = |x| + 5$

Step-by-step explanation:

Since both the original and transformed graph have the same x value for their vertex, there was no horizontal shift. However the vertex shifted up. Specifically it shifted up 5 units. Therefore:

$g(x) = f(x) + 5$

The plus 5 performs a translation up 5 units. Plugging in for f(x) it can be seen that:

$g(x) = |x| + 5$

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3 years ago

25 penalties decreased by 32%?

SCORPION-xisa [38]

Kut 32% off 25

100-32=68% left

=25*0.68

=17

6 0

3 years ago

How to find out the slope

grin007 [14]

To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points .

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2 years ago

What the same as 1.4

liubo4ka [24]

Answer:

$\frac{7}{5}$

Step-by-step explanation:

You can make 1.4 as a fraction, and it will be the same but in a different form.

So, to make a decimal by a fraction, you make 1.4 into a whole number. 1.4 will be 14.

Now, the denominator will be 10.

So, when you simplify $\frac{14}{10}$ , which is $\frac{7}{5}$

3 0

3 years ago

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

2 years ago