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3 years ago

Who is vladimir lenin?

2 answers:

7 0

Vladimir Lenin, also called Vladimir Ilich Lenin, original name Vladimir Ilich Ulyanov (born April 10 [April 22, New Style], 1870, Simbirsk, Russia—died January 21, 1924, Gorki [later Gorki Leninskiye], near Moscow), founder of the Russian Communist Party (Bolsheviks), inspirer and leader of the Bolshevik Revolution ( ...Sep 30, 2016

Goshia [24]3 years ago

5 0

Known by the alias Lenin, a Russian communist revolutionary, politician, and political theorist.

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B5%7D%20%20%2B%20%20%5Cfrac%7B1%7D%7B3%7D%20" id="TexFormula1" title="\frac{3}

Scrat [10]

The answer would be 14/15

Find a common denominator that both of the fractions can use. An easy way to do this is simply multiply the denominators together. So for 3/5 you would multiply by 3 to get the fraction 9/15 and then multiply 1/3 by 5 to get 5/15. Add both 9/15 and 5/15 to then get 14/15.

4 0

3 years ago

Read 2 more answers

Solve for x. Express the answer in simplest form. 2/1 = 8x - 2/ 9

oksano4ka [1.4K]

2=8x-2/9
add 2/9 to both sides
2 and 2/9=8x
2=18/9
2 and 2/9=18/9+2/9=20/9
20/9=8x
multiply both sides by 9
20=72x
divide boht isdes by 72
20/72=x=5/18

4 0

3 years ago

-3+9-27+81-243+729. Using the sigma notation to write the given sum

IgorC [24]

Answer:

546

Step-by-step explanation:

5 0

3 years ago

Can someone please solve this with explanation?

Pepsi [2]

Answer: 67/9 or 7 4/9

Step-by-step explanation:

4 0

3 years ago

Calculus piecewise function.

Kipish [7]

Part A

The notation $\lim_{x \to 2^{+}}f(x)$ means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows $\lim_{x \to 2^{+}}f(x) = 2$

Then for the left hand limit $\lim_{x \to 2^{-}}f(x)$ , we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, $\lim_{x \to 2^{-}}f(x) = 1$

===============================================================

Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.

4 0

1 year ago