(2x² - 5) - (3x² + 4).
Pretend that there is a -1 in front of :(3x² + 4).
(2x^2-5)-1(3x^2+4)
2x^2-5-3x^-4
2x^2-3x^2-5-4
=-x^2-9
Answer: -x^2-9
Answer: -3
Step-by-step explanation:
18/6 = 3
18/-6 = -3
Answer:
The factors are (5x + 3) and (2x + 1)
Step-by-step explanation:
When you need to factor a quadratic, and the coefficient of the x² is not 1, use the slide and divide method.
The general form of a quadratic is ax² + bx + c
Factor: 10x² + 11x + 3
Here a = 10, b = 11, and c = 3
Step 1: Multiply ac, we SLIDE a over to c. Notice the 10 is gone for now..
x² + 11x + 30
Step 2: Factor this (this step will always factor)
x² + 11x + 30 = (x + 5)(x + 6)
So the factors are (x + 5)(x + 6), but we now need to DIVIDE by a, since we multiplied it into c before. We divide the constants in the factors...
(x + 5/10 )(x + 6/10 )
Now reduce the fractions as much as possible...
(x + 1/2 )(x + 3/5)
*If they don't reduce to a whole number, SLIDE the denominator over as a coefficient of x....
(2x + 1)(5x + 3) *2 slide over in front of x, 5 slide over in front of x, the fractions are gone!
These are our factors!
Consider the universal set R,R, define the interval A=[−7,1],A=[−7,1], interval B=(−1,5),B=(−1,5), and CC be the negative real n
AURORKA [14]
Draw a diagram representing the real number line, and the sets A and B.
According to the diagram:
<span>1. A∪B= [-7, 5)
2. </span><span>2. A∩B∩C= (-1, 0)
</span>3. A∩(B∪C)= [-7, 1] <span>∩ (-INFINITY, 5) = [-7, 1]
4. </span>4. A^c ∪ B^c∪ C^c= (not A) ∪ (not B) ∪ (not C) =
(-infinity, -7) ∪ (1, infinity) ∪ (-infinity, -1] ∪ [5, infinity) ∪ [0, infinity)
= (-infinity, -1] ∪ (1, infinity ) ∪ [0, infinity) = (-infinity, -1] ∪ [0, infinity)
5. (C−A)∪(B−C)= "in C but not in A" union "in B but not in C"
=(-infinity, -7) ∪ [0, 5)
Answer: Is this a question? Or statement Yes they can be angles
Step-by-step explanation: