Answer:
1. D
2. B
3. A
Step-by-step explanation:
Question 1:
The pair of <JKL and <LKM can be referred to as linear pairs. They are two adjacent angles that are formed from the intersecting of two lines.
Question 2:
Given that <KLM = x°
<KML = 50°
<JKL = (2x - 15)°
According to the exterior angle theorem, exterior ∠ JKL = <KLM + KML.
2x - 15 = x + 50
Solve for x
2x - x = 15 + 50
x = 65
Therefore, <KLM = 65°
QUESTION 3:
<JKL = 2x - 15
Plug in the value of x
<JKL = 2(65) - 15
= 130 - 15
<JKL = 115°
The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.
Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.
By the fundamental theorem of calculus,
The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so
Answer:
no solution
Step-by-step explanation:
X/3+2=x/3
Subtract x/3 from each side
X/3-x/3+2=x/3-x/3
2 = 0
This is never true so there is no solution
Answer:3 is nonlinear ,1 is function ,domain is the input and the range is the output, mx is the slope and the b is the y intercept
Step-by-step explanation:
I think it’s 20 but I may be wrong
