All Answers 358187 Answers

\(\pi{r^2}=A\)

Lets just say that the radius is 1

So the area would be pi.

If the radius is increased 100%, then the new radius is 2.

Susbtituting 2 into the equation, the new area would be 4pi.

\(\frac{4\pi}{\pi}=4\)

Based on this, can you figure out the answer?

Um... can you upload the image it is blocked on my browser. They categorized it as Illegal ****

x < (-1/2, 2/3)

Prime factorize

72

8 * 9 

2³ * 3²

Factor out a 6²

2 * 6²

Make system of equations

2x + 2y = 2

x + y = 1

Solving, we get x = 1/2 and y = 1/2

yea, you are correct

followup questions:

For what values of a does the equation ax +a^2=0

have a solution in the interval (−1, 2)?

For what values of a does the equation ax +a^2=0

perform for all x from the interval (−1, 2)? 

(1/2, 1/2)    ?

Im not sure but the probability is either 1/6 or 7. Please correct if wrong.

The least common multiple of 12, 18, and 30 is 180.

Use Binary sequences ( a sequence of ones and zeroes) to help visualize.

Someone flips the switch on every single lightbulb

Every lightbulb starts turned on. ON is 1, OFF is 0.

1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

Someone flips the switch on every other lightbulb
Someone flips the switch on every third lightbulb
Someone flips the switch on every fourth lightbulb

This means that LCM of 2, 3, 4,. Which is 12.

A binary sequence of 12 terms would repeat in the a hundred light bulbs.

So lets have 12 light bulbs.

101010101010 Someone flips the switch on every other lightbulb

100011100011 Someone flips the switch on every third lightbulb

100111110010 Someone flips the switch on every fourth lightbulb

There are 7 out of 12 lightbulbs that are turned on.

Now we have a pattern of 100111110010 repeating until the number of terms is 100. We have to find the number of 1s in that 100-length pattern.

Every 12 terms that are seven 1s.

So 12 * 8 = 96

8 * 7 = 56

So out of 96 terms, 56 are 1s.

There are 4 more terms.

Counting up the pattern 100111110010 four terms, we have two 1s.

So 56 + 2 = 58

58 light bulbs are turned on

Say 'x' is the value before it goes up

it goes up 25 %    then it equals   1.25 x

then it goes down to the original value of x

1.25  x   -  (%1.25 x) = x      divide through by x

1.25 - %1.25  = 1

1 - % = 1 / 1.25

% = .2           20 percent  

The answer is (6,0,2).

The answer is sqrt(18).

lol i had a solution ready, then deleted it because I thought it continued to 100. nuuuuuuuu

EDIT: I remade it,

 below is the explanation

Sorry, I tried to edit my answer, but Web2.0 didn't is glitching or something. I can't publish my answer, it just reverts back.

Just look at EP's answer, that is the correct method.

Of course the flipping stops after the foruth bulb. It would be too dark to see anything if more were turnd off. 

x^3 =x^2 +4x

x(x^2-x-4) = 0

x = 0 , 1/2+ - sqrt(17)/2

Three intercepts

Substitute y

So now you have:

x³=x²+4x

0 = x³ - x² - 4x

This is a polynomial with a degree of 3, therefore it must have 3 solutions.

If it has 3 solutions, it has 3 intersections in the graph.

Does the flipping only go to the 4th light bulb?

Or does he keep flipping until he flips all 100 lightbulbs?

Yeah, I just set two equal, no need to overthink it.

Here is my attempt at this. Somebody should check it!
A=Apple, B=Banana, C=Cherry, P=Pina Colada
{A, A, B, B} | {A, A, B, C} | {A, A, B, P} | {A, A, C, C} | {A, A, C, P} | {A, A, P, P} | {A, B, B, B} | {A, B, B, C} | {A, B, B, P} | {A, B, C, C} | {A, B, C, P} | {A, B, P, P} | {A, C, C, C} | {A, C, C, P} | {A, C, P, P} | {A, P, P, P} | {B, B, B, C} | {B, B, B, P} | {B, B, C, C} | {B, B, C, P} | {B, B, P, P} | {B, C, C, C} | {B, C, C, P} | {B, C, P, P} | {B, P, P, P} | {C, C, C, C} | {C, C, C, P} | {C, C, P, P} | {C, P, P, P} | {P, P, P, P} (total: 30)