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2 years ago

Help please it is graphing

Mathematics

1 answer:

Alja [10]2 years ago

4 0

Here are the pairs:

(from the left)

First group: 3rd graph, y = x - 2, table g

Second group: 2nd graph, y = 2x, table b

Third group: 1st graph, y = x + 2, table m

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Please help me!!!!

Free_Kalibri [48]

Answer: (ln10)/6=t

Step-by-step explanation: e is an irrational number, like pi. Therefore the only variable to solve for is "t." Divide both sides by 4, then subtract (1/4) from both sides to get 10. Then take the natural log of both sides because you get rid of e on the left side, and you have the variable by itself. Natural log, or ln, represents loge base e, so dividing both sides by 6, we get t! I hope this helps!

7 0

1 year ago

Read 2 more answers

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

Which quadratic equation has roots - 2 and 1/5 :

Sladkaya [172]

Answer:

$5 {x}^{2} \: + \: 8x \: - 2 \: = \: 0$

Step-by-step explanation:

$sum \: of roots = - 2 + \frac{1}{5} = - \frac{8}{5} \\ product \: of \: roots = - 2 \times \frac{1}{5} = - \frac{2}{5} \\ from \: general \: equation \\ a {x}^{2} + bx + c = 0 \\ {x}^{2} + \frac{8}{5} x \: - \frac{2}{5} = 0 \\ 5 {x}^{2} \: + \: 8x \: - 2 \: = \: 0$

7 0

2 years ago

I need help ASAP giving brainliest!!

Olin [163]

Answer:

isosceles and right

I'm not sure

3 0

2 years ago

Read 2 more answers

13.48x - 200 < 256.12

podryga [215]

Answer:

x < 33.84

Step-by-step explanation:

we have

13.48x-200 < 256.12

Solve for x

Adds 200 both sides

13.48x-200 +200 < 256.12+200

13.48x < 456.12

Divide by 13.48 both sides

13.48x/13.48 < 456.12/13.48

x < 33.84

The solution is the interval ----> (-∞, 33.84)

All real numbers less than 33.84

5 0

2 years ago