<span>dividend x3 - 11x2 + 16x + 84
- divisor * x2 x3 - 7x2
remainder - 4x2 + 16x + 84
- divisor * -4x1 - 4x2 + 28x
remainder - 12x + 84
- divisor * -12x0 - 12x + 84
remainder 0
Quotient : x2-4x-12 Remainder: 0
Step-1 : Multiply the coefficient of the first term by the constant 1 • -12 = -12
Step-2 : Find two factors of -12 whose sum equals the coefficient of the middle term, which is -4 .
-12 + 1 = -11
-6 + 2 = -4 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -6 and 2
x2 - 6x + 2x - 12
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-6)
Add up the last 2 terms, pulling out common factors :
2 • (x-6)
Step-5 : Add up the four terms of step 4 :
(x+2) • (x-6)
Which is the desired factorization
Final result :
(x + 2) • (x - 6) • (x - 7)</span>
The answer to your question is as follows
281.12
I think that the answer is 58,800 watts
The solution of a linear inequality in two variables like Ax + By > C is an ordered pair (x, y) that produces a true statement when the values of x and y are substituted into the inequality.
Example
Is (1, 2) a solution to the inequality
2x+3y>1
2⋅1+3⋅2>?1
2+5>?1
7>1
The graph of an inequality in two variables is the set of points that represents all solutions to the inequality. A linear inequality divides the coordinate plane into two halves by a boundary line where one half represents the solutions of the inequality. The boundary line is dashed for > and < and solid for ≤ and ≥. The half-plane that is a solution to the inequality is usually shaded.
Example
Graph the inequality
y≥−x+1
picture44
For the first problem, the triangle is equilateral. This means that all three sides will have the same length. So you need to find a value for x that makes all three of those statements equivalent. I believe you could find it using a system of equations... but it’s been a while so don’t quote me on that.
For the second, ce and de are equal. Set them equal to each other and solve for x. Then plug that x value back into the side lengths.
