How to solve -2(-4f+5)=-7+2f-9
2 answers:
Answer:
Step-by-step explanation:
First, use PEMDAS to distribute the -2 to the parenthesis.
Combine like terms on the right side.
Now, you may either subtract the f's or the constants. I'll subtract the constants first, meaning I will add 10 to both sides.
Now subtract the 2f from both sides.
Divide both sides by 10.
Simplify.
You might be interested in
So hmm notice the graph below
based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"
now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be
well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either
what's the "p" distance? well, we just found it, is just 1/8
so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8
and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic
based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"
now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be
well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either
what's the "p" distance? well, we just found it, is just 1/8
so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8
and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic
Answer:
No
Reasoning:
If something is a perfect cube, it is able to be put under a cube root () and will result in an integer (a non-decimal number > 0, basically).
So let's calculate , and see if the result is an integer.
= 3.634.......
As you can see, the result is not an integer, therefore 48 is not a perfect cube.