All Answers 352936 Answers

Uhhhh.... what the heck... (i am a girl btw)

Are u ok. That's creepy and please stop harassing people.

Wait, I have one quick question. I do not get your solution. I want to understand the problem so I am wondering if you could tell me a tiny bit more about the step where you have the equation 2x^2 + x - 6 and the +, -, and 0's.

Why is there 2 differnet 0's and +'s?

You were very helpful, thanks for answering my question.

Yeah, it was a snorkeling accident with coral. Thanks so much though!

s                           

\(\text{surgery eh...}\\ x^2 + x + 3 = \left(x+\dfrac 1 2\right)^2 +\dfrac{11}{4} >0\\ 2x^2 + x - 6 = (2x-3 )(x+2)\\ 2x^2 + x - 6 \begin{cases}+ &x<2\\0&x=2\\- &2 < x < \dfrac 3 2\\ 0 &x=\dfrac 3 2\\+&\dfrac 3 2 < x\end{cases}\)

\(+ \div + \Rightarrow +\\ +\div - \Rightarrow -\\ \dfrac{x^2+x+3}{2x^2+x-6}\geq 0 \Rightarrow (-\infty,2]\cup \left[\dfrac 3 2,\infty\right)\)

36 possible rolls...only 3 add up to 10      4 6   5 5   6 4     3/36 = 1/12

No problem. Always glad to help! 

6. This one is fairly nice.

Label the numbers: x-2, x-1, x, x+1, x+2, x+3.

So, we have 6x+3=153, 6x=150, x=25

Thus, their product is \(23*24*25*26*27*28=\boxed{271252800}.\)

tysm!

all correct

#4

\(\text{assume we transfer }n \text{ birds at a time and it was assumed that }\\ \text{transferring }113 \text{ birds would take }t \text{ days. }\\ r \text{ is the common odd number of birds for the last trip}\\ 113 = n(t-1) + r\\ 131 = n(t+2)+r\)

\(\text{subtract 1 from 2}\\ 18=3n\\ n=6\)

\(131 = 6(t+2)+r\\ 119 = 6t + r\\ \text{note that we are told }r < n\\ \left \lfloor \dfrac{119}{6} \right \rfloor= 19\\ 119=19 \cdot 6 + r\\ 119=114+r\\ r=5 \)

These are plug in questions (as far as checking goes) and you are just being lazy.

2)

\(\text{we seek }d \text{ such that}\\ i d = \dfrac{18}{4},~j d = \dfrac{30}{4},~k d = \dfrac{11}{4}~ft,~i,j,k \in \mathbb{Z}\\ LCM(18,30,11) = 990\\ \text{so split up the longest side into }990 \text{ cubes}\\ d = \dfrac{30}{4} \times \dfrac{1}{990} = \dfrac{1}{132}~ft\\ \text{The prism will be split into }\\ \dfrac{\frac{18}{4}}{\frac{1}{132}}\times \dfrac{\frac{30}{4}}{\frac{1}{132}}\times \dfrac{\frac{11}{4}}{\frac{1}{132}}=\\ 595 \times 990 \times 363 = 213465780 \text{ cubes}\)

Here is my take :)

\(\sqrt{0.64}-\sqrt[3]{-0.008}+\sqrt[4]{\frac{4}{20.25}}+\sqrt[5]{16\div\frac{1}{2}}\\ =0.8-\sqrt[3]{-1*8*0.001}+\sqrt[4]{\frac{4}{2025\div100}}+\sqrt[5]{32}\\ =0.8-(-0.2)+\sqrt[4]{\frac{4*100}{2025}}+2\\ =0.8+0.2+\sqrt[2]{\frac{2*10}{45}}+2\\ =3+\sqrt[2]{\frac{2*10}{5*9}}\\ =3+\sqrt[2]{\frac{2*2}{9}}\\ =3+\frac{2}{3}\\ =3\frac{2}{3}\\\)

Set h >=6   and solve for t

6>= -4.9 t^2 + 14 t - 0.4

0 >= -4.9 t^2 + 14 t - 6.4     Use Quadratic Formula or graphing to find   t = .5714  and   2.286  sec

So between    (.5714, 2.286)    the cannonball is = to or higher than 6 m