BuilderBoi

Add \(5x^2\) to both sides: \(-3 = 13x^2 - 3\)

Add 3 to both sides: \(0 = 13x^2\)

Divide both sides by 13: \(0 = x^2\)

Take the square roots of both sides: \(x = \color{brown}\boxed{0}\)

160 and 40.

The set would contain 1 number that is divisible by 7, and as many "sets" of numbers that leave a remainder of 1, 2, and 3 when divided by 7. 

The smallest "set" of numbers would be 1, 2, and 3, and the largest "set" of numbers would be 2017, 2018, and 2019. 

The smallest number is \(7 \times 0 + 1\) and the largest number is \(7 \times 288 + 1 \), so there are \(288 - 0 + 1 = 289\) "sets" of numbers. 

So, set S can have at most \(289 \times 3 + 1 = \color{brown}\boxed{868}\) elements.  

Let the base be b. Off the bat, we know that \(b>8\).

If we added everything, it would look something like this:

        66

        87 

+      85

        48 

\(\overline{\color{white}{wwwwww}}\)

       132

Adding up the ones digit gives us 26, so we have \(26 \mod b = 2\)

The only options for b are b = 12 or 24. 

If b = 12, the sum of the ones digit would be \(22_{12}\), so we could carry the 2 to the tens column. Adding everything up would give us \({24}_{12}\), so our answer would be \(242_{12}\).

This means that the base is \(\color{brown}\boxed{24}\).

(Add it up to see for yourself.)

that's not a problem...

Because the graph is continuous when x = 2, \(x- 5 = ax+ 3\). Subbing in what we know for x, we get \(a = -3\)

Now, do the same thing when x = -2.

Just make a list of the numbers and cross out the ones that don't work. 

The discriminant \(b^2 - 4ac\) must be 0. So we have the equation \(196 -4c = 0\), and we need to solve for c. 

Can you take it from here?

Melody also answered this here: https://web2.0calc.com/questions/binomial-probability_6

Instead of finding the probability that two or more babies will speak tomorrow, we will find the probability that 0 or 1 babies speak tomorrow and subtract that from 1. 

The probability that none of the babies speak tomorrow is \(({3 \over 4})^5 = {243 \over 1024}\)

The probability that exactly 1 baby speaks tomorrow is \({1 \over 4} \times ({3 \over 4})^4 \times {5 \choose 1} = {405 \over 1024}\) (1 baby has to speak, the others must not, and there are 5 ways to choose the baby that does speak)

So, the probability that at least 2 babies speak tomorrow is \(1 - {243 \over 1024} - {405 \over 1024} = {376 \over 1024} = \color{brown}\boxed{47 \over 128}\)