All categories

4 years ago

If the function y=x^5 is transformed to y=x^5+3 what’s the statement

Mathematics

1 answer:

Anastaziya [24]4 years ago

7 0

I dont know what you mean by the question but according to me.

If y=x^5

y=x^5+3

Then y+3=x^5+3

Answered by Gauthmath must click thanks and mark brainliest

You might be interested in

Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^

Snezhnost [94]

I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to $x^2+2y^2+3z^2=96$ .

Set $f(x,y,z)=xyz$ and $g(x,y,z)=x^2+2y^2+3z^2-96$ , so that the Lagrangian is

$L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)$

Take the partial derivatives and set them equal to zero.

$\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}$

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

$\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}$

So by adding the first three equations together, you end up with

$\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0$

and the fourth equation allows you to write

$\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda$

Now, substituting this into the first three equations in the most recent system yields

$\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}$

So we found a grand total of 8 possible critical points. Evaluating $f(x,y,z)=xyz$ at each of these points, you find that $f(x,y,z)$ attains a maximum value of $\dfrac{128}{\sqrt3}$ whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of $-\dfrac{128}{\sqrt3}$ whenever exactly one or all of the critical points' coordinates are negative.

6 0

3 years ago

Helpppppppppppp!<br> this is an exam

kari74 [83]

Answer:

84.7 but don't cheat on exams from now on ;)

8 0

3 years ago

Read 2 more answers

A bag contains 240 marbles that are either red, blue, or green. The ratio of red to blue to green marbles is 5:2:1. If one-third

Andrej [43]

The fraction of the remaining marbles in the bag will be blue is 6/17.

<h3>What fraction of the remaining marbles in the bag will be blue?</h3>

The first step is to determine the initial number of marbles in the bag:

Initial number of red marbles in the bag : (5/8) x 240 = 150

Initial number of blue marbles in the bag : (2/8) x 240 = 60

Initial number of green marbles in the bag : 240 - 150 - 60 = 30

Number of red marbles remaining after 1/3 is removed = (1 - 1/3) x 150

2/3 x 150 = 100

Number of green marbles remaining after 2/3 is removed = (1 - 2/3) x 30

1/3 x 30 = 10

Total number of marbles now in the bag : 100 + 10 + 60 = 170

Fraction of blue marbles = 60 / 170 = 6/17

To learn more about ratios, please check: brainly.com/question/9194979

#SPJ1

7 0

2 years ago

Can you construct a triangle that has side lengths 2 yd, 9 yd, and 10 yd?

yan [13]

Yes

The rule to check if a triplet of numbers can represent the sides of a triangle is the following: the sum of the length of two sides must exceed the third one.

This happens for all possible couples in your case:

$2+9=11>10$