All Answers 358233 Answers

a) Yes, as long as a and b are allowed to be complex numbers.  Consider the simple parabola

\(y = x^2+1 \\ \text{with complex roots } \pm i \\ \text{this can be written as }\\ (x-i)(x+i) = x^2 +i x - i x +(i)(-i) = x^2+1\)

b) No.  Consider a general cubic polynomial

\(y = a x^3 + b x^2 + c x + d\)

\(\text{as }x \to -\infty, \text{the }x^3 \text{ term dominates everything and }y \to -\infty \\ \text{as }x \to +\infty, \text{ likewise the }x^3 \text{ term dominates and }y \to +\infty \\ \text{somewhere along the way it's got to cross the x axis and at that point }\\ y=0 \text{ at a real value of x}\)

All the numbers you have listed are divisible by 4 !!. The rule for divisibility by 4 is this:

If the first TWO DIGITS of any number are EVENLY divisible by 4, then the whole number, no matter how big, is divisible EVENLY by 4. Example:3769321324. Since the last two digits, or 24, are EVENLY divisible by 4, then that whole number is EVENLY divisible by 4. And that is it.

See the answer here: https://web2.0calc.com/questions/in-the-arrangement-below-each-number-is-the-non-negative

Help anybody!

Thank you! I did not notice those mistakes.

you left out the lines A'C' and AC

the line A'D' is supposed to be solid

the line AD is supposed to be solid

the lines A'E' and  D'E' are supposed to be only dashed

AE and DE are supposed to be dashed only

since this is a 90 degree triangle we can use trig functions

\(\sin(35^\circ)=\dfrac{x}{15}\\ x = 15 \sin(35^\circ)~cm\)

I'll let you plug and chug those numbers

I would have to go with Axl Rose.

Well, there is no exact year for when the pagans started to celebrate Halloween (at least not that I could find or remember) but there is a time frame. Our version of Halloween and festivities are taken from Samhain which is what the pagan and Celtic called it. Our historical records show that this holiday dates back to about 2,000 years ago. That would place around AD 18 obviously. There are 206 bones in our bodies. 18 + 206 = 224. 224 to the 6th power is 122978496247489 or 1.5957918e+28 since I got two different answers somehow. I think that I am extremely wrong though. But I did have fun doing this even if I am wrong.

Thank you

Welcome back, just in time for the spook

Yes when it was a pagan holiday, and thank you

 Howdy! Welcome Back.

When you say "start" do you mean when it was originally a Celtic and Pagan holiday, when it started in the U.S., when it started in Europe, or when Christians decided to make it their own?

I love your costume idea!

Suppose a railroad rail is

44

kilometers and it expands on a hot day by

1111

centimeters in length. Approximately how many meters would the center of the rail rise above the​ ground?

c) 6 feet is 72 inches.  So total up the bins 71-72 and less.

10+7+6+3 = 26

d) 6' 2" is 74 inches.  So only count the last bin which has 2 in it.

a) \(\left( \begin{array}{cc} 4 & \{2,3,6,6,7,7,8,9,9\} \\ 5 & \{0,1,1,1,1,1,2,2,4,4,4,4,4,5,5,5,5,6,6,6,7,7,7,7,8\} \\ 6 & \{0,1,1,1,2,4,4,5,8,9\} \\ \end{array} \right)\)

b) just count up the leafs of 5.  There are 25 of them

c) just look at the largest leaf of the largest stem.  You correctly got 69.

6 are 18 years old so if you add all thier ages together you get 18+18+18+18+18+18=6*18= 108

this about this and also that

the mean is the average and the average is the sum of everybodies ages divided by how many people that there are.

Now can you give it a go ?

\(\mathbb{Z} \subset \mathbb{Q} \\ \\ \text{simply says that the integers are a subset of the rationals}\)

\(\text{The strike zone looks to be }18 \times 24 = 432~sq~in \text{ in area}\\ \text{each of the triangles is }\dfrac{6^2}{2} = 18~sq~in~\text{in area} \\ \text{Thus the probability of hitting any of those triangles is }\\ p=\dfrac{18}{432} = \dfrac{1}{24}\)

\(\text{The throws are independent so we just multiply the individual probabilities together to}\\ \text{get the probability of all 4 events occurring}\\ p = \left(\dfrac{1}{24}\right)^4 = \dfrac{1}{331776}\)

First , let's do the right multplctn

15x^2 -25x - 12x + 20       -2x+3     now collect 'like' terms

15x^2 -39x + 23            Is that what you need? 

a) \(p = \dfrac{1}{13} \dfrac{1}{2} = \dfrac{1}{26}\)

b) \(p = \dfrac{1}{2}\dfrac{1}{4} = \dfrac{1}{8}\)

c) \(p = \dfrac{1}{13} \dfrac{3}{13} = \dfrac{3}{169}\)

d) \(p = \dfrac{1}{52}\dfrac{1}{13} = \dfrac{1}{676}\)