Divide. Show your work. (6x^4-5x^3+6x^2)/2x^2

1 answer:

6 0

We can do like we did before, distribute the denominator,

$\bf \cfrac{6x^4-5x^3+6x^2}{2x^2}\implies \cfrac{6x^4}{2x^2}-\cfrac{5x^3}{2x^2}+\cfrac{6x^2}{2x^2}\implies \cfrac{6}{2}\cdot \cfrac{x^4}{x^2}-\cfrac{5}{2}\cdot \cfrac{x^3}{x^2}+\cfrac{6}{2}\cdot \cfrac{x^2}{x^2}
\\\\\\
3 x^4x^{-2}-\cfrac{5}{2}x^3x^{-2}+3x^2x^{-2}\implies 3x^{4-2}-\cfrac{5}{2}x^{3-2}+3x^{2-2}
\\\\\\
3x^2-\cfrac{5}{2}x+3x^0\implies 3x^2-\cfrac{5}{2}x+3$

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For her presentation on the Wonders of the World, Mary baked a square pyramid-shaped cake as pictured below. The slant height of

mihalych1998 [28]

Answer:

$V=196in^3$

Step-by-step explanation:

The volume of a pyramid is:

$V=\frac{A_{b}h}{3}$

where $A_{b}$ is the area of the base and $h$ is the height (the perpendicular measurement between base and highest point, not the slant height)

Since the base is a square, the area is given by:

$A_{b}=l^2$

where $l$ is the length of the side: $l=8in$ , thus:

$A_{b}=(8in)^2\\A_{b}=64in^2$

Now we need to find the height, for this we use the right triangle that forms with half of a square side (8in/2 = 4in), the slant height (10in), and the height.

In this right triangle, the slant height is the hypotenuse, the leg 1 is the unknown height, and leg 2 is half of the square side.

Using pythagoras:

$hypotenuse^2=leg1^2+leg2^2$