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4 years ago

14

Two college friends rent an apartment. They must pay the landlord two months' rent and $500 security deposit when they sign the

lease. The total amount they pay the landlord is $2800. How much does it cost to rent the apartment for one month?

1 answer:

lianna [129]4 years ago

6 0

Answer:it costs $1150 to rent the apartment for one month.

Step-by-step explanation:

Let x represent the cost to rent the apartment for one month.

They must pay the landlord two months' rent and $500 security deposit when they sign the lease. This means that the security deposit is constant and independent of the number of months that they want to pay for.

The total amount they pay the landlord is $2800. Therefore

2x + 500 = 2800

2x = 2800 - 500 = 2300

x = 2300/2 = 1150

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No

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4 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

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3 years ago

4. Which rotation about the origin of line segment MN results in line segment M’N’?

olganol [36]

For this, let's examine the options, if it is A, with a rotation of 45 ° clockwise, you would have the following

If it is C or D, with a 180 ° rotation in a clockwise or counterclockwise direction, the segment would remain the same.

Finally, let's examine the option, B with a 90 ° rotation clockwise, you have the following.

Therefore, the correct answer is B. A 90° clockwise rotation.

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1 year ago

You paint four walls. Each wall is a rectangle with a length of 20 feet and a height of 12 feet. One gallon of paint covers abou

Naddik [55]

Answer:

3 gallons

Step-by-step explanation:

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960 divided into 320 = 3

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3 years ago

Use theorem 7. 4. 2 to evaluate the given laplace transform. do not evaluate the convolution integral before transforming. (writ

irga5000 [103]

With convolution theorem the equation is proved.

According to the statement

we have given that the equation and we have to evaluate with the convolution theorem.

Then for this purpose, we know that the

A convolution integral is an integral that expresses the amount of overlap of one function as it is shifted over another function.

And the given equation is solved with this given integral.

So, According to this theorem the equation becomes the

$\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{\frac{s+1}{(s+1)^2+1}}{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).$

Then after solving, it become and with theorem it says that the

$\mathscr{L} \left( \int_{0}^{t} f(\tau) d\tau \right) = \frac{\mathscr{L} ( f(\tau))}{s} .$

Hence by this way the given equation with convolution theorem is proved.

So, With convolution theorem the equation is proved.

Learn more about convolution theorem here

brainly.com/question/15409558

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3 0

2 years ago