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Elena-2011 [213]
3 years ago
6

What is the value of x in the solution of the system of equations below?8x-y=20Y=3x​

Mathematics
2 answers:
SashulF [63]3 years ago
5 0

Answer:

32-12=20 or x=4

Step-by-step explanation:

This is basically saying 8x-3x=20

so first you want to do 8x-3x to end up with 5x

then from there you have the equation..

5x=20

in order to figure out what x is you need to divide 20 by five so you just have the x and get rid of the 5

so 20 divided by 5 = 4

this means x=4 but you need to plug it into the equation

8x= 8 times 4 = 32   and   3x= 3 times 4 = 12

so 32-12=20

tatyana61 [14]3 years ago
4 0

Answer:

x=4

where y =3x

8x-3x=20

5x=20

5x/5=20/5

x=4

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The number of daily sales of a product was found to be given by S = 600xe−x2 + 600 x days after the start of an advertising camp
Elena-2011 [213]

Answer:

a. 20520

b. 12600

Step-by-step explanation:

Given the function S = 600xe^-x² + 600.

a. To find the average, we have to find the definite integral of the function because average, as it is known, is the sum of data points divided by the size of its dataset, this can be used for discrete data. Integral formula is just the continuous form of average, so we are using integral because we were given an interval of x= 0 to X = 30.

Let's integrate 100xe^x² first. Let –x² = u, this means –2xdx = du and we have dx = –du/2x. Also, when x = 0, u = –(0)² = –0 and when x = 30, u = –(30)² = –900. When we make our substitutions we have:

–600(xe^udu)/2x = –600(e^udu)/2 upon integrating that we have –600(e^u)/2. Applying our interval we have

–600[(e^900)/2 – (e^0)/2] ≈ – [– 3.7 – (1/2)] = –600 (–4.2) = 600 x 4.2 = 2520

Now let's integrate 600, with the interval x = 0 to x = 30 (we are using this interval here because the substitution we made didn't affect this).

We have, upon integrating:

600x.

Substituting our intervals we have:

600(30 – 0) = 600 x 30 = 18000.

Adding that up we have: 2520 + 18000 = 20520.

b. The same method is needed, just difference of interval.

Therefore, after integrating the first component with intervals u = 900 to u = 2500 (from x² = u) we have:

–600[(e^2500)/2 – (e^900)/2] ≈ –600[2.7 – 3.7] = –600(–1) = 600.

Then for the second component:

600x using x = 30 to x = 50

600(50 – 30) = 600 x 20 = 12000.

Adding that up we have:

12000 + 600 = 12600.

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