To answer this question, we need to find the midpoint of \(\displaystyle \overline{CD}\).

To find how far the midpoint of a line is from each end, we use the following equation:

\(\displaystyle midpoint \:distance=\frac{x_{2}-x_{1}}{2}, \frac{y_{2}-y_{1}}{2}\)

\(\displaystyle x_{2}\) and \(\displaystyle y_{2}\) are taken from the \(\displaystyle x,y\) value of the second point and \(\displaystyle x_{1}\) and \(\displaystyle y_{1}\) are taken from the \(\displaystyle x,y\) value of the first point. Therefore, for this data:

\(\displaystyle midpoint \:distance=\frac{x_{2}-x_{1}}{2}, \frac{y_{2}-y_{1}}{2}=\frac{7-(-2)}{2}, \frac{1-(-1)}{2}\)

We can then solve:

\(\displaystyle \frac{7-(-2)}{2}, \frac{1-(-1)}{2}=\frac{9}{2},\frac{2}{2}=4.5,1\)

Therefore, our midpoint is \(\displaystyle 4.5\) units between each endpoint's \(\displaystyle x\) value and \(\displaystyle 1\) unit between each endpoint's \(\displaystyle y\) value. To find out the location of the midpoint, we subtract the midpoint distance from the \(\displaystyle x_{2},y_{2}\) point. (In this case it's the point \(\displaystyle 7,1\).) Therefore:

\(\displaystyle 7-4.5= 2.5\)

\(\displaystyle 1-1=0\)

So the midpoint is located at \(\displaystyle 2.5,0\)

The question asked us what the \(\displaystyle x\)-coordinate of this point was. Therefore, our answer is \(\displaystyle 2.5\).

The first step is to find the y-coordinates for the two points we are using. To do this we plug our x-values into the equation. Where \(\displaystyle x=1\), we get \(\displaystyle \frac{4}{3}-4=-\frac{8}{3}\), giving us the point \(\displaystyle \left(1,-\frac{8}{3}\right)\). Where \(\displaystyle x=4\), we get \(\displaystyle \frac{16}{3}-4=\frac{4}{3}\), giving us the point \(\displaystyle \left(4,\frac{4}{3}\right)\).

We can now use the distance formula: \(\displaystyle d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}\).

Plugging in our points gives us \(\displaystyle d=\sqrt{(1-4)^2+\left(-\frac{8}{3}-\frac{4}{3}\right)^2}=\sqrt{(-3)^2+(-4)^2}=\sqrt{25}=5\)

To answer this question, we must put the equation into slope-intercept form, meaning we must solve for \(\displaystyle y\). Slope-intercept form follows the format \(\displaystyle y = mx + b\) where \(\displaystyle m\) is the slope and \(\displaystyle b\) is the intercept.

Therefore, we must solve the equation so that \(\displaystyle y\) is by itself. First we add \(\displaystyle 4\) to both sides so that we can start to get \(\displaystyle 2y\) by itself:

\(\displaystyle 10x+2y-4 +4=0+4\rightarrow 10x+2y=4\)

Then, we must subtract \(\displaystyle 10x\) from both sides:

\(\displaystyle 10x+2y -10x=4-10x\rightarrow 2y=-10x+4\)

We then must divide each side by \(\displaystyle 2\)

\(\displaystyle \frac{2y}{2}=\frac{-10x+4}{2}\rightarrow y=-5x+2\)

Therefore, the slope-intercept form of the original equation is \(\displaystyle y=-5x+2\).

The volume of a triangular prism can be simply stated as V = A*L, where A is the area of the triangular face and L is the length. We have the volume already and we need to figure out the area of the triangle from the side length.

The area of a triangle is b*h/2, and we are given the base/side length: 3 in. We also have a formula for the height of an equilateral triangle, \(\displaystyle b\frac{\sqrt{3}}{2}\). Calculating the area of the triangle we get 2.5981 in^2, so now we just have to plug our numbers into the volume formula.

\(\displaystyle V = A\cdot L\Rightarrow 32 in^3 = 2.5981 in^2 \cdot L\)

\(\displaystyle L = \frac {32}{2.5981} in = 12.32 in\)

And that is the final answer.

First, find the volume of the pool. For a rectangular prism, the formula for the volume is the following:

\(\displaystyle \text{Volume}=\text{length}\times \text{width}\times \text{height}\)

For the swimming pool,

\(\displaystyle \text{Volume}=10 \times 25\times 3=750\) cubic meters

Now, because the hose only fills up \(\displaystyle 5\) cubic meters per hour, divide the total volume by \(\displaystyle 5\) to find how long it will take for the pool to fill.

\(\displaystyle 750\div5=150\)

It will take the pool \(\displaystyle 150\) hours to fill by hose.

First, find the volume of the cake. For a rectangular prism,

\(\displaystyle \text{Volume}=\text{length}\times\text{width}\times\text{height}\)

\(\displaystyle \text{Volume}=15\times12\times4=720\)

Next, find the volume of each individual slice.

\(\displaystyle \text{Volume}=3\times2\times4=24\)

Now, divide the volume of the entire cake by the volume of the slice to get how many pieces of cake Matt can cut.

\(\displaystyle 720\div24=30\)

First, find the side lengths of the cube.

Recall that the surface area of the cube is given by the following equation:

\(\displaystyle \text{Surface Area}=6s^2\), where \(\displaystyle s\) is the length of a side.

Plugging in the surface area given by the equation, we can then find the side length of the cube.

\(\displaystyle 150=6s^2\)

\(\displaystyle s^2=25\)

\(\displaystyle s=5\)

Now, recall that the volume of a cube is given by the following equation:

\(\displaystyle \text{Volume}=s^3\)

\(\displaystyle \text{Volume}=5^3=125\)

The volume of a right triangular prism is given by the following equation:

\(\displaystyle \text{Volume}=\text{Area of the Base}\times \text{height}\)

Now, for the given question, the height is \(\displaystyle 12\).

\(\displaystyle 120=\text{Area of the base}\times 12\)

\(\displaystyle \text{Area of the base}=10\)

Since the area of the base is a right triangle, we can plug in the given values to find \(\displaystyle x\).

\(\displaystyle \text{Area of Triangle}=\frac{\text{base}\times\text{height}}{2}\)

\(\displaystyle 10=\frac{5x}{2}\)

\(\displaystyle 5x=20\)

\(\displaystyle x=4\)

The volume of a right triangular prism is given by the following equation:

\(\displaystyle \text{Volume}=\text{Area of the Base}\times \text{height}\)

\(\displaystyle \text{Area of Base}=\frac{5\times6}{2}=15\)

\(\displaystyle \text{Volume}=15\times 8=120\)

For a rectangular prism, the formula for the volume is the following:

\(\displaystyle \text{Volume}=\text{length}\times \text{width}\times \text{height}\)

Now, we know that the height is twice its width. We can rewrite that as:

\(\displaystyle h=2w\)

\(\displaystyle w=\frac{h}{2}\)

We also know that the height is half its length. That can be written as:

\(\displaystyle h=\frac{1}{2}l\)

\(\displaystyle l=2h\)

Now, we can plug in the values of the length, width, and height in terms of height to find the height.

\(\displaystyle 64=(2h)(\frac{h}{2})(h)\)

\(\displaystyle 64=h^3\)

\(\displaystyle h=4\)

The question wants to find the length of the box. Plug in the value of the height in the earlier equation we wrote earlier to represent the relatioinship between the height and the length.

\(\displaystyle l=2h=2(4)=8\)