Answer: (2, -10)
Down and left are negative right and up are positive.
Answer:
1
Step-by-step explanation:
using PEDMAS(parentheses, exponents, division, multiply, addition, subtraction) to solve the problem
First we multiply (4)(-3)
(4)(−3)−5−3(−6)
Then multiply -3(-6)
(-12)-5-3(-6)
=-12-5+18
subtract 12-5
-12-5+18
=-17+18
Adding -17+18 gives us 1.
Therefore, (4)(−3)−5−3(−6) is 1.
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


The y-intercept is always the number at the end of a standard slope equation, so in your case, it’s - 3
Step-by-step explanation:
(o, -2) is the correct answer because in -90°
P [x,y] =P' [y,-x ]
hope it is helpful to you