Step 1: Find f'(x):
f'(x) = -6x^2 + 6x
Step 2: Evaluate f'(2) to find the slope of the tangent line at x=2:
f'(2) = -6(2)^2 + 6(2) = -24 + 12 = -12
Step 3: Find f(2), so you have a point on y=f(x):
f(2) = -2·(2)^3 + 3·(2)^2 = -16 + 12 = -4
So, you have the point (2,-4) and the slope of -12.
Step 4: Find the equation of your tangent line:
Using point-slope form you'd have: y + 4 = -12 (x - 2)
That is the equation of the tangent line.
If your teacher is picky and wants slope-intercept, solve that for y to get:
y = -12 x + 20
D. NI = AC is the needed information to prove ΔINF = ΔCAT by the ASA Postulate.
For the triangles to be congruent by ASA, the measurements of two angles and one side must be proven congruent. Since two sets of congruent angles are already given, one side must also be congruent.
When proving congruency using ASA, the congruent parts of the triangles must be in this order: Angle, Side, Angle. So, you have to the side that is between the two sets of congruent angles.
Answer:
y=1
x=7/4
Step-by-step explanation:
4x + 5y = 12
3x + 4y = 9.25
4x + 5y = 12
12x + 16y =37
-12x -15 y =-36
12x + 16y =37
y=1
x=7/4
Basically you do fractions side by side. so, 6/3=15/h and then cross multiply to get your answer so that would give you 3h=80, then divide 80÷3
