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3 years ago

9

find a function whose second derivative is y''=12x-2 at each point (x,y) on its graph and y=-x+5 is tangent to the graph at the

point corresponding to x=1

1 answer:

pogonyaev3 years ago

3 0

Start by integrating y'' twice to get function y(x).

$y' = \int (12x - 2) dx = 6x^2 -2x + c \\ \\ y = \int (6x^2 -2x+c )dx= 2x^3-x^2 +cx+d$

Next find value of coefficients 'c' and 'd' using the given tangent line
The slope of tangent line is equal to y' when x=1.

$6x^2 -2x+c = -1, x = 1 \\ \\ 6-2+c = -1 \\ \\ c = -5$

The tangent line intersects the curve y(x) when x = 1.
If x = 1, then y = 4 (From y =-x+5)

$y = 2x^3 -x^2-5x +d , x = 1, y = 4 \\ \\ 2-1-5+d = 4 \\ \\ d = 8$

Final Answer:
$y = 2x^3 -x^2 -5x+8$

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fomenos

Answer:

D. (-2,-5), (0, -7), (1, -4)

Step-by-step explanation:

From Function Theory we must remember that range of a function is the set of images related to elements of the domain. In this case, we must find the image of each of the three elements that forms the domain of $f(x, y) = (y+2, -x-4)$ , which are: $A(x,y) = (1, -4)$ , $B(x,y) = (3,-2)$ and $C(x,y) = (0,-1)$ .

Then we proceed to find all elements of range:

$A(x,y) = (1, -4)$

$f(1,-4) = (-4+2,-1-4)$

$f(1,-4) = (-2,-5)$

$B(x,y) = (3,-2)$

$f(3,-2) = (-2+2,-3-4)$

$f(3,-2) = (0,-7)$

$C(x,y) = (0,-1)$

$f(0,-1) = (-1+2,0-4)$

$f(0,-1) = (1,-4)$

Which corresponds to option D.

3 0

3 years ago

PLEASE HELP ASAP What is the m and b for this graph??

suter [353]

Answer:

20. 9. 23 12.131. 32.2.3..2.3.3.1.3 13.2.3 2

3.30am. 3.2 36.2. 1.. 3.2.3 3.2.3.3 3.3. 33. 43 3.3...3.34 2. 3 4 3.4 2.4.4. and the. 4. I. 4.4 4 3.34. is. that. 3.. I think 3 and. But. You. You 3.2 42. 44 3.4.3 2.3 4 3.3.3 4 3 3 4 2. 34.3.3 3 3.2.3 3.2.3.3. You. can. I. am. 34 3 3.43.2 34 3

3 0

2 years ago

Read 2 more answers

Studentka2010 [4]

Answer:

The area of the shaded portion of the figure is $9.1\ cm^2$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The shaded area is equal to the area of the square less the area not shaded.

There are 4 "not shaded" regions.

step 1

Find the area of square ABCD

The area of square is equal to

$A=b^2$

where

b is the length side of the square

we have

$b=4\ cm$

substitute

$A=4^2=16\ cm^2$

step 2

We can find the area of 2 "not shaded" regions by calculating the area of the square less two semi-circles (one circle):

The area of circle is equal to

$A=\pi r^{2}$

The diameter of the circle is equal to the length side of the square

so

$r=\frac{b}{2}=\frac{4}{2}=2\ cm$ ---> radius is half the diameter

substitute

$A=\pi (2)^{2}$

$A=4\pi\ cm^2$

Therefore, the area of 2 "not-shaded" regions is:

$A=(16-4\pi) \ cm^2$

and the area of 4 "not-shaded" regions is:

$A=2(16-4\pi)=(32-8\pi)\ cm^2$

step 3

Find the area of the shaded region

Remember that the area of the shaded region is the area of the square less 4 "not shaded" regions:

so

$A=16-(32-8\pi)=(8\pi-16)\ cm^2$

---> exact value

assume

$\pi =3.14$

substitute

$A=(8(3.14)-16)=9.1\ cm^2$

8 0

3 years ago

Did I do this correctly?

Rina8888 [55]

Yes!
- you applied the distributive property
- combined like terms
- and answered the rest right!

8 0

3 years ago

The ratio of dogs to cats at a local animal shelter is 13/25. Which statement must be true?

Ivan

<em>Option:</em>

<em>a. for every 13 dogs, there are 25 cats. </em>

<em>b. there are twice as many cats as dogs at the shelter. </em>

<em>c. the shelter has more dogs than cats. </em>

<em>d. there are 25 animals in the shelter.</em>

Answer:

a. for every 13 dogs, there are 25 cats.

<em></em>

Step-by-step explanation:

Represent Dogs with D and Cats with C;

So;

$\frac{D}{C} = \frac{13}{25}$

By comparison:

$D=13$

$C = 25$

This implies that:

For every 13 D, there are 25 C

Hence: Option A is true

6 0

2 years ago