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2 years ago

Identify the solution set of 6 ln e = eln 2x {2} {3} {6}

Mathematics

2 answers:

Sladkaya [172]2 years ago

8 0

Answer:

The solution set is {3}

Step-by-step explanation:

I am assuming the correctly formatted equation looks like this:

$6\cdot \ln e = e^{\ln 2x}$

Given that, the following simplifications on both sides of this equation can be made and x can be solved for:

$6\cdot \ln e = e^{\ln 2x}\\6\cdot 1= 2x\\3 = x$

(recall that ln(e)=1 and e^{ln(x)}=x)

The equation has exactly one solution, namely 3.

aleksley [76]2 years ago

3 0

Answer:

Step-by-step explanation:

in edge

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Ella takes 1 3/5 hours to complete a painting. Holly takes 3/4 of the time that Ella takes. How long will Holly take to complete

Misha Larkins [42]

1. you have to figure out what 3/4 of Ella's time is.
$1 \frac{3}{5} = \frac{8}{5}$
(You find that out by multiplying 5x1. Then add 3.)
2. Convert the fractions into decimals. (This will help make it easier when dividing.)
$\frac{8}{5} = 8 \div 5 = 1.6$
$\frac{3}{4} = 3 \div 4 = .75$
3. Next you are going to divide 1.6 by .75
$1.6 \div .75 = 2.13$
4. Now multiple 2.13 by 5.
$2.13 \times 5 = 10.65$
Answer is 10

3 0

2 years ago

Which relationships have the same constant of proportionality between yyy and xxx as the equation y=\dfrac{5}{2}xy= 2 5 xy, eq

LUCKY_DIMON [66]

Answer:

$8y = 20x$

Step-by-step explanation:

Given

$y=\dfrac{5}{2}x$

Required

Determine equivalent expressions

A. 5y = 2x

$y=\dfrac{5}{2}x$

Multiply through by 5

$5 * y=\dfrac{5}{2}x * 5$

$5y=\dfrac{25}{2}x$

<em>This is not equivalent</em>

B. 8y = 20x

$y=\dfrac{5}{2}x$

Multiply both sides by 8

$8 *y=\dfrac{5}{2}x * 8$

$8y=\dfrac{5 * 8}{2}x$

$8y=\dfrac{40}{2}x$

$8y = 20x$

<em>The other options are not clear.</em>

However, another possible relationship is when

$y=\dfrac{5}{2}x$ is cross multiplied

This gives:

$2y = 5x$

5 0

2 years 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 $\frac{1}{3^3}= \frac{1}{27}$