Area = length *width
A = x²
3x² = (x+10)(x+12)
3x² = x²+10x+12x+120
3x² = x²+22x+120
3x²-3x² = x²+22x+120-3x²
0 = -2x²+22x+120
Factor to solve for x:
-2(x²-11x-60)
(x+4)(x-15)
x = -4 & x = 15 but we can't have a negative length so we eliminate 4. The length of the square is 15 cm.
For this case we must find the solution set of the given inequalities:
Inequality 1:
Applying distributive property on the left side of inequality:
Subtracting 3 from both sides of the inequality:
Dividing by 6 on both sides of the inequality:
Thus, the solution is given by all the values of "x" greater than 3.
Inequality 2:
Subtracting 3x from both sides of the inequality:
Subtracting 3 from both sides of the inequality:
Thus, the solution is given by all values of x less than 4.
The solution set is given by the union of the two solutions, that is, all real numbers.
Answer:
All real numbers
Answer:
I believe A is the answer
<span>The coterminal angles of angle A are given by adding k times 360° to it, where k is an integer, positive or negative.
Similarly, all angles that are coterminal with 141° are given by
141° + k * 360° where k is a positive or negative integer.
That is 141° + 360°, 141 + 2*360° , 141+3*360° and so on
and 141° - 360°, 141° -2*360°, 141° - 3*360° and so on.
9720)</span>
Answer:
1
Explanation:
−((10)(2)−(2)(8)+(5)(−1))
= −(20−(2)(8)+(5)(−1))
= −(20−16+(5)(−1))
= −(4+(5)(−1))
= −(4+−5)
= −(−1)
= 1
- Mordancy.
