Help meee I don't know what to do anymore--

Mathematics

1 answer:

balandron [24]2 years ago

5 0

Answer:

It is increasing by 40 and the first game will 40 the next is 80 the next is 120 and it keeps going on

Step-by-step explanation:

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A rectangular box with a square base is open at the top and has a volume of 12 cubic feet. Each side of the base has a length of

Stolb23 [73]

We know that the side lengths of the square base are: x * x. The volume is 12, so for now, let's say that y is the other side length. Then, x * x * y = 12. We can solve for y: y = 12/x^2. Now, we find the surface area of the 5 sides.

Four of the sides have the same area: x * (12/x^2) = 12/x, so we multiply this by 4: 48/x.

The last side is the base: x * x = x^2.

We add 48/x to x^2:

x^2 + 48/x

So, the answer is the fourth choice, (d).

4 0

3 years ago

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.