-72-4x^2+8x^3-36x/x-3

-4(18+x^2-2x^3+9x)/x-3

-4(-2x^3+x^2+9x+18)/x-3

-4(-2x^2x(x-3)-5x x(x-3)-6(x-3) )/x-3

-4 x(-(x-3) ) x (2x^2+5x+6)/x-3

-4 x (-1) x (2x^2 +5x+6)

8x^2+20x+24

7 0

3 years ago

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At a lunch stand, each hamburger has 50 more calories than each order of fries. If 2 hamburgers and 3 orders of fries have a tot

Vedmedyk [2.9K]

Answer: 960 calories

Step-by-step explanation:

Given

Each hamburger has 50 more calories than each order of fries

Suppose each order of fries has $x$ calories

so, hamburger has $x+50$ calories

There are total of 1700 calories available. Thus, we can write

$\Rightarrow 2(x+50)+3x=1700\\\Rightarrow 2x+3x=1700-100\\\Rightarrow 5x=1600\\\Rightarrow x=320$

So, each order of fries has $320$ calories

For 3 orders, it is $320\times 3=960$ calories

7 0

3 years ago