Jesse rotated the triangle 90° clockwise about the origin to create figure A'B'C'. What are the coordinates of the vertices of the figure A'B'C' after the rotation? * Captionless Image A. A’(-9, 0); B’(-9, 7); C’(-1, 0) B. A’(0, -9); B’(7, -9); C’(0, -1) C. A’(0, 9); B’(7, 9); C’(0, 1) D. A’(-9, 0); B’(-9, -7); C’(-1, 0)

you never say the total so we cant find the percentage used for petrol

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

step-by-step explanation:

because it just is

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.