Answer:
x = 3y + 5...........1
4x - 2y = - 10...,..2
substituting value of x in equation 2
4(3y+5)-2y=-10
12y+20-2y=-10
10y=-10-20
y=-30/10=-3 is answer
substituting value of y in equation 1
x=3×-3+5=-9+5=-4 is answer
Answer:9
Step-by-step explanation:
A..Just plus it if not wrong
Answer:
Step-by-step explanation:
Given: f(x)=5x^3-51x^2+77x+100/x^2-11x+24
Please use parentheses to eliminate any ambiguity:
f(x) = (5x^3-51x^2+77x+100) / (x^2 - 11x + 24)
or (better yet):
5x^3-51x^2+77x+100
f(x) = ---------------------------------
x^2 - 11x + 24
The vertical asymptotes here are at the zeros of the denominator:
x^2 - 11x + 24 = 0, This quadratic equation has coefficients a = 1, b = -11 and c = 24. Thus, its roots (zeros) are:
-(-11) ± √( 121 - 4(1)(24) )
x = -------------------------------------
2(1)
or:
11 ± √( 25 )
x = --------------------
2
or: x = 8 and x = 3
The vertical asymptotes are x = 8 and x = 3.
If we attempt to divide x^2 - 11x + 24 into 5x^3 - 51x^2 + 77x + 100, we see that the first term of the quotient is 5x. As x increases or decreases without bound, 5x goes to either ∞ or -∞, so we conclude that there is no horiz. asymptote. Continuing this division results in:
5x + 4 + a fraction
This represents the slant asymptote, y = 5x + 4
well, let's take a look at the table
is g(x) increasing from -2 to -1? hmmm from when x = -2 g(x) = 2, then when x = -1, g(x) = -3, hold the mayo!! it went from 2 to -3, that's just going down, so, nope is not.
is itdecreasing from -1 to 0? hmmm x = -1, g(x) = -3, as x = 0, g(x) = 2, wait a second!!! it went from -3 to 2, it really went up, so nope, is not decreasing on that interval.
is it increasing from 0 to 1? let's see, as x = 0, g(x) = 2, as x = 1, g(x) = 17, woahhh!!, it went from 2 and rocketted to 17, yeap, it is increasing as x = 0 to x = 1.