Answer: Solving for f. Want to solve for x instead?
1 Remove parentheses.
f\times -2fx=3{x}^{2}-8x+7f×−2fx=3x
2
−8x+7
2 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x
a
x
b
=x
a+b
.
-{f}^{2}\times 2x=3{x}^{2}-8x+7−f
2
×2x=3x
2
−8x+7
3 Regroup terms.
-2{f}^{2}x=3{x}^{2}-8x+7−2f
2
x=3x
2
−8x+7
4 Divide both sides by -2−2.
{f}^{2}x=-\frac{3{x}^{2}-8x+7}{2}f
2
x=−
2
3x
2
−8x+7
5 Divide both sides by xx.
{f}^{2}=-\frac{\frac{3{x}^{2}-8x+7}{2}}{x}f
2
=−
x
2
3x
2
−8x+7
6 Simplify \frac{\frac{3{x}^{2}-8x+7}{2}}{x}
x
2
3x
2
−8x+7
to \frac{3{x}^{2}-8x+7}{2x}
2x
3x
2
−8x+7
.
{f}^{2}=-\frac{3{x}^{2}-8x+7}{2x}f
2
=−
2x
3x
2
−8x+7
7 Take the square root of both sides.
f=\pm \sqrt{-\frac{3{x}^{2}-8x+7}{2x}}f=±√
−
2x
3x
2
−8x+7
8 Simplify \sqrt{-\frac{3{x}^{2}-8x+7}{2x}}√
−
2x
3x
2
−8x+7
to \sqrt{\frac{3{x}^{2}-8x+7}{2x}}\imath√
2x
3x
2
−8x+7
ı.
f=\pm \sqrt{\frac{3{x}^{2}-8x+7}{2x}}\imathf=±√
2x
3x
2
−8x+7
ı
9 Regroup terms.
f=\pm \imath \sqrt{\frac{3{x}^{2}-8x+7}{2x}}f=±ı√
2x
3x
2
−8x+7
Done- :)
f=±ı√ 2x 3x 2 −8x+7
Step-by-step explanation
What this setup essentially represents is “How many 8s can we take away from 32 before hitting 0?” Which in turn can be reframed as “How many 8s fit into 32?” This can be captured in the expression 32 / 8. As we can see from the problem, we can take away 4 8’s before hitting 0, so that gives us the equation 32 / 8 = 4
A geometric solid is another name for the traditional 3-dimensional object with 3 dimensions: width, length, and height.
Some examples of a geometric solid include cubes and triangular prisms.
Answer:
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Step-by-step explanation:
