What is the cofficient in the expression 8x^2+1?

Please enter the missing number: 4, 5, 8, 17, 44, ?

Eddi Din [679]

Answer:

81

Step-by-step explanation:

from 4 to 5,the difference is 1 ,and when you multiply by 3, then add 3 to 5 ,it gives you 8 ,to get 17,I multiple 3by 3 and add the answer to 8 ,and the same continues upto when you get the final answer which is 81.

difference is just multiplying the previous difference with 3 and add it with the previous number .

3 0

3 years ago

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,

vichka [17]

Answer:

A(-1,0) is a local maximum point.

B(-1,0) is a saddle point

C(3,0) is a saddle point

D(3,2) is a local minimum point.

Step-by-step explanation:

The given function is

$f(x,y)=x^3+y^3-3x^2-3y^2-9x$

The first partial derivative with respect to x is

$f_x=3x^2-6x-9$

The first partial derivative with respect to y is

$f_y=3y^2-6y$

We now set each equation to zero to obtain the system of equations;

$3x^2-6x-9=0$

$3y^2-6y=0$

Solving the two equations simultaneously, gives;

$x=-1,x=3$ and $y=0,y=2$

The critical points are

A(-1,0), B(-1,2),C(3,0),and D(3,2).

Now, we need to calculate the discriminant,

$D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$

But, we would have to calculate the second partial derivatives first.

$f_{xx}=6x-6$

$f_{yy}=6y-6$

$f_{xy}=0$

$\Rightarrow D=(6x-6)(6y-6)-0^2$

$\Rightarrow D=(6x-6)(6y-6)$

At A(-1,0),

$D=(6(-1)-6)(6(0)-6)=72\:>\:0$ and $f_{xx}=6(-1)-6=-18\:$

Hence A(-1,0) is a local maximum point.

See graph

At B(-1,2);

$D=(6(-1)-6)(6(2)-6)=-72\:$

Hence, B(-1,0) is neither a local maximum or a local minimum point.

This is a saddle point.

At C(3,0)

$D=(6(3)-6)(6(0)-6)=-72\:$

Hence, C(3,0) is neither a local minimum or maximum point. It is a saddle point.

At D(3,2),

$D=(6(3)-6)(6(2)-6)=72\:>\:0$ and $f_{xx}=6(3)-6=12\:>\:0$

Hence D(3,2) is a local minimum point.

See graph in attachment.

3 0

3 years ago

There are "n" delicious sweets in the bag. You know that 6 of them are orange and the rest are yellow. Hannah takes a random swe

nikitadnepr [17]

At the point when Hannah takes her first sweet from the sack, there is a 6/n chance it is orange.
This is because that there are 6 orange desserts and n desserts altogether.
When Hannah takes out her second sweet, there is a 5/(n-1) chance that it is orange.
This is because there are just 5 orange desserts let alone for an aggregate of n-1 desserts.
The possibility of getting two orange desserts in succession is the main likelihood increased by the second one: 6/n x 5/n–1
The question lets us know that the shot of Hannah getting two orange desserts is 1/3.
So: 6/n x 5/n–1 = 1/3
Now, rearrange this problem.
(6x5)/n(n-1) = 1/3
This gets to be:
30/(n² – n) = 1/3
Times by 3 on both sides:
90/(n² – n) = 1
What's more, doing likewise with (n² – n):
So (n² – n) = 90
Our answer is: n² – n – 90 = 0

3 0

3 years ago

What is the solution to the equation x + 5.7 = 6.1?

Ann [662]

The solution is X=0.4

4 0

3 years ago

Read 2 more answers

Use the figure below and 2-3 sentences to explain the following:

nydimaria [60]

Answer:

See below

Step-by-step explanation:

The side opposite angle E is side e (DF)

The side adjacent to angle E is side d (EF)

The hypotenuse is side f (ED)

5 0

3 years ago