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3 years ago

5

Lisa had a set of 12 toys cars.Then she gave her sister 3.What fraction of Lisa's set did Lisa give her sister. A 3/4 B 1/4 C 2/

3 D 1/3

2 answers:

love history [14]3 years ago

4 0

Lisa fraction is 1/4,but it could be a

nexus9112 [7]3 years ago

3 0

B 1/4 because 12/3 equals 4 so she only gave 1 of the 4 she had so the answer would be 1/4

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I REALLY NEED HELP WITH ALL OF THESE!!

const2013 [10]

Answer:

1. negative 2. zero 3 undefined 4 positive

Step-by-step explanation:

4 0

2 years ago

What is the perfect square number is the base factor is 11?

Svetlanka [38]

Answer:

121

Step-by-step explanation:

we know that

A <u><em>perfect square</em></u> is an integer that is the square of an integer.

In this problem

The integer or base factor is equal to 11

therefore

squared the integer to find out the perfect square

$11^2=121$

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3 years ago

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

2 years ago

If bc is the diameter of the circle below, what is the area of the circle

dangina [55]

Answer:

option a.

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2 years ago

Read 2 more answers

In each box is a step in the process of solving a max/min problem like the one in Question 8. Number these steps in order (1-8)

Dmitriy789 [7]

Answer:

The steps are numbered below

Step-by-step explanation:

To solve a maximum/minimum problem, the steps are as follows.

1. Make a drawing.

2. Assign variables to quantities that change.

3. Identify and write down a formula for the quantity that is being optimized.

4. Identify the endpoints, that is, the domain of the function being optimized.

5. Identify the constraint equation.

6. Use the constraint equation to write a new formula for the quantity being optimized that is a function of one variable.

7. Find the derivative and then the critical points of the function being optimized.

8. Evaluate the y-values of the critical points and endpoints by plugging them into the function being optimized. The largest y- value is the global maximum, and the smallest y-value is the global minimum.

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2 years ago