MATHEMATICS HIGH SCHOOL

Ex 2.11
20) A curve y''=12x-24 and a stationary point at (1,4). evaluate y when x=2.

Answers

Answer 1
Answer: So, dy/dx=0 at the point (1, 4) - that is where x=1 and y=4.

\int { 12x-24dx } \n \n =\frac { 12{ x }^( 2 ) }{ 2 } -24x+C\n \n =6{ x }^( 2 )-24x+C

\n \n \therefore \quad { f }^( ' )\left( x \right) =6{ x }^( 2 )-24x+C

But when x=1, f'(x)=0, therefore:

0=6-24+C\n \n 0=-18+C\n \n \therefore \quad C=18

\n \n \therefore \quad { f }^( ' )\left( x \right) =6{ x }^( 2 )-24x+18

Now:

\int { 6{ x }^( 2 ) } -24x+18dx\n \n =\frac { 6{ x }^( 3 ) }{ 3 } -\frac { 24{ x }^( 2 ) }{ 2 } +18x+C

=2{ x }^( 3 )-12{ x }^( 2 )+18x+C\n \n \therefore \quad f\left( x \right) =2{ x }^( 3 )-12{ x }^( 2 )+18x+C

Now when x=1, y=4:

4=2-12+18+C\n \n 4=8+C\n \n C=4-8\n \n C=-4

\n \n \therefore \quad f\left( x \right) =2{ x }^( 3 )-12{ x }^( 2 )+18x-4

Now when x=2,

f\left( x \right) =2\cdot { 2 }^( 3 )-12\cdot { 2 }^( 2 )+18\cdot 2-4\n \n =16-48+36-4\n \n =0

So when x=2, y=0.
Answer 2
Answer: y''=12x-24\ny'=\int 12x-24\, dx\ny'=6x^2-24x+C\n\n0=6\cdot1^2-24\cdot1+C\n0=6-24+C\nC=18\ny'=6x^2-24x+18\n\ny=\int 6x^2-24x+18\, dx\ny=2x^3-12x^2+18x+C\n\n4=2\cdot1^3-12\cdot1^2+18\cdot1+C\n4=2-12+18+C\nC=-4\n\n 2x^3-12x^2+18x-4

y(2)=2\cdot2^3-12\cdot2^2+18\cdot2-4\ny(2)=16-48+36-4\n\boxed{y(2)=0}

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Find the area and perimeter of an isosceles trapezoid with legs 25cm and bases 16 cm and 30 cm.

Answers

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F (x) = cx + 3In the function above, c is a constant. If f (1) = 1, what is the value of f (-4) ?

Answers

If f(1) = 1 that means c is -2 because when you substitute x = 1 and when c = -2, you'll get f(1) = -2(1)+3 = -2+3 = 1

So the constant c is -2 and that means the value of f(-4) is

f(-4)=-2(-4)+3=8+3=11

Thus, the value of f(-4) is 11.

About why does "c" have to be -2, here's the proof.

 Given y = f(x), therefore.

y=cx+3

Since when x = 1 and y = 1, then Substitute x = 1 and y = 1, thus 1 = c+3

1 - 3 = c and therefore, -2 = c #

So this is the proof that c = -2

55.57What is the equation of the line that is parallel to the
given line and passes through the point (12, -2)?
Oy=-x +10
Oy=-x + 12
Oy=-x- 10
Oy=x - 12

Answers

The equation of the line that is parallel to the given line and passes through the point (12, -2) is y = 5/6(x)-12

What is Slope of Line?

The slope of the line is the ratio of the rise to the run, or rise divided by the run. It describes the steepness of line in the coordinate plane.

The slope intercept form of a line is y=mx+b, where m is slope and b is the y intercept.

The slope of line passing through two points (x₁, y₁) and (x₂, y₂) is

m=y₂-y₁/x₂-x₁

The slope of line passing through the points (12, 6) and (0, -4)

Slope = -4-6/0-12

=-10/-12

=5/6

Now let us find y intercept passing through (12, -2)

-2=5/6(12)+b

-2=10+b

-12=b

Hence,  the equation of the line that is parallel to the given line and passes through the point (12, -2) is y = 5/6(x)-12

To learn more on slope of line click:

brainly.com/question/16180119

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Answer:10

Step-by-step explanation:and a slope 2/12 roof slope means your roof has 2 inches of vertical drop for every 12 inches of horizontal distance.Aug

1. If angle A and angle B are vertical angles and mA = 7x-5 and mB = 4x + 10, what is the value of x

Answers

Answer:

175/11 or 165 and 10/11

Step-by-step explanation:

7x-5+4x+10=180

11x+5=180

11x=175

x=175/11

Write an equation of the line below.

Answers

Answer:

y=1x+2

Step-by-step explanation:

the gradient is equal to 1 and the c in the question is 2 because the y-intercept represents c.

Gradient= (y2-y1)/(x2-x1)

Y is eleven times the value of x.

The equation is y =____

Answers

Answer:

Your answer will be y = 11x

The same thing as multiplying 11 by x

Or you could say y = 11*x

Hope this comes to help someone! :)
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