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3 years ago

When x2+3x-4 is subtracted from x (x2+3x-2) what is the difference

2 answers:

5 0

I'm assuming this is x^2 + 3x - 4 and x(x^2 + 3x - 2)

1.) First distribute x(x^2 + 3x - 2) to get x^3 + 3x^2 - 2x.

2.) Because you are subtracting all the terms from x^3 + 3x^2 - 2x, it's the same thing as distributing -1 to x^2 + 3x - 4 and then adding it to x^3 + 3x^2 - 2x.

3.) -1(x^2 + 3x - 4) = -x^2 - 3x + 4

4.) Add (x^3 + 3x^2 - 2x) + (-x^2 - 3x + 4)

5.) x^3 + 2x^2 - 5x + 4 is your final answer.

anygoal [31]3 years ago

3 0

Assuming that the problem is (x^2+3x-4) - x(x^2+3x-2) then this would be simple.. simplifying XD
here is the work for you:
(x^2+3x-4) - x(x^2+3x-2)
(x^2+3x-4) - (x^3+3x^2-2x)
Final Answer: -x^3-2x^2+5x-4

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3 years ago

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2 years ago

Find the limit, if it exists. (If an answer does not exist, enter DNE.) lim (x, y) → (0, 0) x4 − 34y2 x2 + 17y2

HACTEHA [7]

Answer:

Step-by-step explanation:

Given the limit of the function $\lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}$ , to find the limit, the following steps must be taken.

Step 1: Substitute the limit at x = 0 and y = 0 into the function

$= \lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}\\= \frac{0^4-34(0)^2}{0^2+17(0)^2}\\= \frac{0}{0} (indeterminate)$

Step 2: Substitute y = mx int o the function and simplify

$= \lim_{(x,mx) \to (0,0)} \frac{x^4-34(mx)^2}{x^2+17(mx)^2}\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^4-34m^2x^2}{x^2+17m^2x^2}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2(x^2-34m^2)}{x^2(1+17m^2)}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2-34m^2}{1+17m^2}\\$

$= \frac{0^2-34m^2}{1+17m^2}\\\\= \frac{34m^2}{1+17m^2}\\\\$

<em>Since there are still variable 'm' in the resulting function, this shows that the limit of the function does not exist, Hence, the function DNE</em>

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3 years ago

If p:q = 1/3:2 and p:r = 1/2:3/4 calculate the ratio p:q:r

kirza4 [7]

Answer:

p:q=1/3:2=1:6=3:18

p:r=3/4:1/2=3:2

p:q:r=3:18:2

i hope its help to you ✌

5 0

2 years ago