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3 years ago

What are the domain and the range of this function?

Mathematics

2 answers:

Pani-rosa [81]3 years ago

5 0

Answer:

(1,5) (-5,-4)

Step-by-step explanation:

Alexandra [31]3 years ago

3 0

Answer:

Step-by-step explanation:

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Factor 13c(a+4)−5b(a+4)

alisha [4.7K]

Answer:

(13c-5b)(a+4)

Step-by-step explanation:

Since both (13c) and (-5b) are being multiplied by (a+4), we can factor both of them out and put them in a bracket.

13c(a+4)−5b(a+4) = (13c-5b)(a+4)

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3 years ago

Find the slope of the line -2,-6 and -4,8

Karolina [17]

Answer:

dhkmbhuigago katangina mo

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3 years ago

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choose the point-slope form of the equation below that represents the line that passes through the points (-3,2) and (2,1)

UkoKoshka [18]

The answer should be: (2 - 1) = m(-3 - 2)

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3 years ago

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Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to $(xyz)^2$ then it means that it is positive i.e $f(x,y,z)\geq 0$ .

Consider the function $F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)$

We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)

$\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0$

$\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0$

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

$-2y\lambda = -2z\lambda$ . If lambda is not zero, then y =z. But, since $-2y\lambda=0$ and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set $y^2+z^2=1$ .

If both x,y are zero, then we have the solution set $z^2=1$ . We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

$\lambda = y^2z^2 = x^2z^2$ which implies $x^2=y^2$

Also, from the first and third equation we have that

$\lambda = y^2z^2 = x^2y^2$ which implies $x^2=z^2$

So, in this case, replacing this in the restriction we have $3z^2=1$ , which gives as another solution set. On this set, we have $x^2=y^2=z^2=\frac{1}{3}$ . Over this solution set, we have that the value of our function is $\frac{1}{3^3}= \frac{1}{27}$

4 0

3 years ago

What is the product of 187 x 70?

Semenov [28]

ANSWER: 13090

EXPLANATION: okay so basically whenever you see the word "product" it means that you just have to multiply. So their asking what is 187 x 70 and well, it's 13090
hopefully this helps you in the future!! :D

6 0

2 years ago