C*h=area/2
type i am going to say acute triangle
IDK if i have to explain again but
A. (-infinity, infinity)
B. (-infinity, -3]
C. (-infinity, -4)u(-4, 0)u(0, infinity)
those u's are the logic symbol
2. domain: [-6, -1)u(-1, 3]u [3, -infinity)
range: (-infinity, 4]
3a. 8x^2 + 17
3b. 32x^2 + 80x + 53
3c. I cant see the multiplication symbol
The first step is to write this equation into general form. The
general form of an equation is:
ax^2 + bx + c = 0
To make this equation to general form, you have to simplify
the equation first.
2/3(x-4) (x+5) = 1
2/3 (x^2 + 5x – 4x – 20) = 1
2/3(x^2 + x -20) = 1
2/3x^2 + 2/3x – 40/3 = 1
2/3x^2 + 2/3x – 40/3 – 1 = 0
2/3x^2 +2/3x – 43/3 = 0
Therefore, a = 2/3 ; b = 2/3 ; c = -43/3
Answer:
yes
Step-by-step explanation:
Answer:
a = -1
Step-by-step explanation:
-6(3a+4) = 3a-3
-18a-24 = 3a-3
-24+3 = 3a+18a
21a = -21
a = -21 ÷ 21
a = -1
