The straight line with equation y = 11x - 28
is a tangent to the curve
y =  {x}^(3)  - a {x}^(2)  + bx - 1
at the point (3,5). Find A, the coefficient of
{x}^(2)

Answers

Answer 1
Answer:

Answer:

a=3

Step-by-step explanation:

The equation of the curve is

y=x^3-ax^2+bx-1.

The gradient function is given by

(dy)/(dx)=3x^2-2ax+b

The gradient of this curve at the point (3,5) is equal to the gradient of the tangent y=11x-28 which is 11.

\Rightarrow 11=3(3)^2-2a(3)+b

\Rightarrow 11=27-6a+b

\Rightarrow b-6a=-16...(1)

The given point, (3,5) also satisfies the equation of the curve.

\Rightarrow 3^3-a(3)^2+b(3)-1=5

\Rightarrow 27-9a+3b-1=5

\Rightarrow3b-9a=-21

\Rightarrow b-3a=-7...(2)

Equation (2) minus equation (1) gives

-3a+6a=-7+16

3a=9

This implies that;

a=3


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What is 18/42 in the simplest form?

Answers

(18)/(42)=(18:2)/(42:2)=(9)/(21)=(9:3)/(21:3)=\boxed{(3)/(7)}
18 | 2 
9 | 3
3 | 3
1

42 | 2
21 | 3
7 | 7


2,3
GCF (18,42) = 2*3 = 6
(18)/(42) =  (18:6)/(42:6) =  (3)/(7)

Can someone help me with this math question?

Answers

Answer:

y = x + 2

Step-by-step explanation:

the formula for this basic equation is

y = mx + b

m = 5/5   (slope)

b = 2   (y-intercept)

5/5 simplified is just 1

so if we put it in the equation you get y = 1x + 2

but 1x is the same as just saying x

so

y = x + 2

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Answers

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Answers

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