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3 years ago

If the length of the side of a square is 3×-y what is the area of the square in terms of x and y

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

5 0

(3x -y)(3x-y )= 9x^2 - 6xy + y^2

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HELP! ITS EASY BUT I CANT DO IT!! Find the volume of a rectangular prism with length = 8.4mm, width = 3.5mm and height = 5.6mm.

Inga [223]

Answer:

164.64

Step-by-step explanation:

you just have to multiply

8.4 x 3.5= 29.4 then you do 29.4 x 5.6 = 164.64

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2 years ago

The table shown below lists the time, in days, that it takes for

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Answer:gotchu g there it is

Step-by-step explanation:

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1 year ago

Jasmine is buying beans. She bought r pounds of red beans that cost $3 per pound and b pounds of black beans that cost $2 per po

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D and B are the answers because they are the variables.

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3 years ago

Sam buys 3 different types of fruit in 3/4 pounds basket. Maria has 1 1/4 pound fruit. How many pounds of fruit do they have alt

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3 years ago

Read 2 more answers

A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c

slava [35]

Answer:

$I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

$\frac{dI}{dt}=a*I*(1000-I)$

Rearranging and integrating

$\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t$

At the initial breakout (t=0) there was one trooper infected (I=1)

$-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755$

In two days (t=2) there were 5 troopers infected

$-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725$

Rearranging, we can model the number of infected troops (I) as

$-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

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2 years ago