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Hello : let : y = f(x) so : y' = f'(x)
calculate : f'(x) and : f'(-1) given f(-1) = 2
by derivate : <span>4x3+2y2−11=4xy−x
12x² +4yy' = 4y +4xy' - 1
4yy' - 4xy' = - 12x² +4y -1
y' ( 4y - 4x) = 4y -12x² -1 ...(*)
if x = -1 y = 2 subsct in : (*)
y' (4(2)-4(-1)) = 4(2)-12(-1)² - 1
2y' = - 9
y' = - 9/2 = f'(- 1) ( the slope of the tangent )
</span>of the tangent line to the curve <span>at the point (−1,2) is :
y - 2 = (-9/2)(x +1)</span>
Answer:
m=5
Step-by-step explanation:
multiply 3 with the numbers in the parentheses (2m + 5) when you have 6m + 15 you will plug it into -10=6m+15+5 add your commons which will be 15 + 5 which will be 20 so the equation will be -10=6m+20, now cross your 6m with your -10 which will be -10 + 6m = 20, now cross over your -10 with 20, now your equation will be 6m=20 + 10 so now you will get 6m = 30 now divide both sides by 6 and m=5
