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2 years ago

PLEASE PLEASE PLEASE HELP I WILL GIVE XTRA POINTS AND BRAINALIST TO FIRST PERSON

Mathematics

1 answer:

victus00 [196]2 years ago

6 0

Answer:

the answer is A because it has all numbers less than 15

Step-by-step explanation:

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It is 8 <br> kilometers from Lucy's house to the nearest mailbox. How far is it in meters?

allochka39001 [22]

Answer:

8000

Step-by-step explanation:

multiply the length value by 1000

3 0

2 years ago

1. Name the angle below in four ways. Then classify it as acute, right, obruse, or straight.

Yanka [14]

Answer:

Ok so I know the angle is acute

Step-by-step explanation:

because the angle looks smaller than a 90 degree angle...

7 0

2 years ago

only 6 of the 75 trees in the park are at leat 30 feet tall. what percent of the trees are over 30 feet?

ipn [44]

Answer:

8 %

Step-by-step explanation:

To find the percent of trees that are over 30 ft tall, take the number of trees that are over 30 ft tall over the number of tree

Percent = trees over 30 ft tall/ total number of trees

= 6/75

=.08

Change this to a percent by multiply by 100%

.08 * 100%

8 %

4 0

3 years ago

Identify the domain and range to the following relations and state whether or not the relations are functions. State why or why

Vladimir79 [104]

It’s a function because it passes the vertical line test

6 0

2 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

3 years ago