-6Y-3=3-6Y what the answer

1 answer:

7 0

The answer is the number that 'Y' must be in order for
the equation to be true.
That number is called the 'solution' to the equation.
Here's one way to go about finding it:

You said that - 6y - 3 = 3 - 6y

Add 6y to each side: - 3 = 3

Let's stop right here.
Is there any value of 'Y' that can make -3 = 3 ?
I don't think so.
Your equation has NO SOLUTION.

There is NO NUMBER that you can write in place of 'y'
in your equation
- 6y - 3 = 3 - 6y
and have it work out to be true.

That's OK. There are lots of equations that have no solution.

You might be interested in

Emily deposits $14,500 each year into a retirement account with a 7% simple interest rate. If she deposits the same amount each

soldier1979 [14.2K]

Hello! I believe she would have 58,000 at the end of her fourth year. Hope this helps. :)

6 0

3 years ago

Read 2 more answers

What is mpo?<br> 128°<br> O 1730<br> O 1920<br> O 256°

Paladinen [302]

Answer:

it's the 2nd one because

Step-by-step explanation:

have a nice day

8 0

3 years ago

PLZ HELP 50 POINTS AND BRAINLIEST PLZZZ HELP TIMED!!!

k0ka [10]

Answer:

i think its C

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

How do you do this question?

Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

$\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}$

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

$V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\$

$\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)$

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.