I think its 1/10 miles/hour
The formula in solving the area of a square is Area = a² where "a" is for the length of the side. The area formula in solving a cube is Area = 6a² where "a" is for the length of its side.
Area of square = a²
64 = a²
a = 8 units
Area of cube = 6a²
64= 6a²
a = 3.27 units
The difference of side of the square and side of the cube is shown below:
Difference = 8 - 3.27
Difference = 4.73 units.
The answer is 4.73 units.
<span>x=6</span>, <span>x=−5</span> or <span>x=9</span>
Explanation:
<span><span>f<span>(x)</span></span>=<span>(x−6)</span><span>(x+5)</span><span>(x−9)</span></span>
If all of the linear factors are non-zero, then so is their product <span>f<span>(x)</span></span>.
If any of the linear factors is zero, then so is their product <span>f<span>(x)</span></span>.
Answer:
Solution : (15, - 11)
Step-by-step explanation:
We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )
Row Echelon Form :

Step # 1 : Swap the first and second matrix rows,

Step # 2 : Cancel leading coefficient in row 2 through
,

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

As you can see our solution is x = 15, y = - 11 or (15, - 11).
Domain is the entire span left to right (on the x-axis) that the graph is on. Since the graph goes from x=-4 and ends at x=4, the domain would be from -4 to 4. The circle at -4 is open, so it does not include the point at -4, just everything leading up to it. So, the domain would be

The range is similar, it is the entire span that the graph goes up and down (on the y-axis). The graph starts at the bottom at y=-2, and ends at y=5. The bottom point (4,-2) is closed, so the graph includes that point, and the top point (-4,5) is open and doesn't include the point. Therefore, the range would be
