All Answers 359226 Answers

A (0,0)

B (b,c)

C (b+ sqrt(bsq + csq), c)

D (sqrt(bsq + csq), 0)

First......let's consider the slope of AC, one of the diagonals....we have

 ([b + √b^2 + c^2] - 0 ) / {c - 0)  =   ([b + √b^2 + c^2] / c  

Next, let's consider the slope of the other diagonal, BD.....we have

[ 0 - c ] / (  ([b + √b^2 + c^2 ] - b ) =  -c / ([b + √b^2 + c^2]

And since the slopes are negative reciprocals, the diagonals are perpendicular to each other

√5cos(2x + 0.464)  =  1      let 2x + 0.464 = θ   divide both sides by √5

cos θ  = 1/√5    using the cosine inverse, we have

cos-1(1/√5) = 1.107  and 5.176 = θ   

So

2x + 0.464  = 1.107     subtract 0.464 from  both sides

2x = .643      divide both sides by 2

x = .3215  or about .322 rounded

And

2x + 0.464  = 5.176

2x = 4.712

x = 2.356 or about 2.36 rounded 

The probability of throwing at least one 4 on a pair of dice is 1 in 6, so the probability of sucess is 1/6. The probability of faliure is 1-1/6 which is 5/6. As it is after 7 rolls, there are 7 trials in the experiment.

Putting these values into an equation for binomial distribution, nCr*P^(r)*Q^(n-r), where P is prob of sucess, q is prob of failure, r is number of sucesses and n is the number of trials, we get...

$${\left({\frac{{\mathtt{7}}{!}}{{\mathtt{1}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}^{\left({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{1}}\right)} = {\mathtt{0.390\: \!714\: \!306\: \!127\: \!114\: \!8}}$$

as the probability of getting a 4 in one roll.

This is very likely wrong.

Anyone better at maths should check this first, since I'm very very unconfident in my binomial skills.

The equation tells you. You subtract a term to make it vanish from a particular side of the equation, usually as part of your manouvering to isolate a variable by itself. In the process of making it disappear from one side, you inevitably cause it to appear on the other side.

Just make sure you are even-handed, what you do to one side must also be done to the other. You can divide everything on one side by some number, as long as you divide all on the other side by that same number. (Just make sure you don't try to do any division by zero!)

Since both are conveniently $${{\mathtt{x10}}}^{-{\mathtt{11}}}$$, we can just divide 2.5 by 1.5

$${\frac{{\mathtt{2.5}}}{{\mathtt{1.5}}}} = {\frac{{\mathtt{5}}}{{\mathtt{3}}}} = {\mathtt{1.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

So a hydrogen atom is 1.6 times larger than a mercury atom.

To first find the volume of the sphere, you need to first find the radius. To find this, you take the circumference of 29.5 inches and divide by pi to get the diameter.

$${\mathtt{C}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{D}}$$

$${\frac{{\mathtt{29.5}}}{{\mathtt{3.14}}}} = {\frac{{\mathtt{1\,475}}}{{\mathtt{157}}}} = {\mathtt{9.394\: \!904\: \!458\: \!598\: \!726\: \!1}}$$

Then divide the diameter by 2 to get the radius.

$${\frac{\left({\frac{{\mathtt{29.5}}}{{\mathtt{3.14}}}}\right)}{{\mathtt{2}}}} = {\frac{{\mathtt{1\,475}}}{{\mathtt{314}}}} = {\mathtt{4.697\: \!452\: \!229\: \!299\: \!363\: \!1}}$$

So we get a radius of about 4.70.

Then we use the formula for the volume of a sphere which is 

$${\mathtt{V}} = {\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{3}}}$$

Substituting in R for 4.69........, we get

$${\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{3.14}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1\,475}}}{{\mathtt{314}}}}\right)}^{{\mathtt{3}}} = {\mathtt{433.965\: \!796\: \!448\: \!807\: \!93}}$$

Rounded to the nearest inch, we get 434 inches cubed.

In any case, thousands of years. But yes anon is right. this isn't web2.0socialstudies.com

It is impossible to know all of the prime numbers as they go on forever.

Actually i'm not sure because i'm tired Lol

Yes you are right AmandXo

Well, study island really isn't the best idea since it gets hacks all the time. Plus anyone can just google the answers. On another note, maybe you should ask this in a chat room.

Since there are 13 diamonds and 52 cards, the probability that the first card is a diamond is 13/52.

Now, there are only 51 cards remaining and 13 of these cards are spades, so the probability that the second card is a spade is 13/51.

Combining these, the probability that the first card is a diamond AND the second card is a spade is:

     13/52 x 12/51.

Multiply these together to get the final probability.

13 diamonds, so probability of 1st card being a diamond is 13/52 = 1/4

Given this, there are 51 cards left, 13 of which are spades, so probability that second card is now a spade = 13/51

Combined probability = (1/4)*(13/51) = 13/204 ≈ 0.064

.

It must stop somewhere (i.e. total stopping probability = 1) so, 

p_A + p_B + p_C + p_D = 1

3/8  +  1/4  + 2p_C = 1   (because p_D = p_C)

5/8 + 2p_C = 1

2p_C = 3/8

p_C = 3/16

.

(4 + 5)3 = 4(3) + 5(3)

Use the distributive property of multiplication, and multiply

12 + 14 = 12 + 15

26 = 27

Those don't equal each other, so I usually put a slash through the =

prob 1st die odd and 2nd die even = (2/3)*(2/3) = 4/9

prob 1st die even and 2nd die odd = (1/3)*(1/3) = 1/9

So probability that sum is odd = 4/9 + 1/9 = 5/9

.

19, but next time use the calculator located in the home page :)

ooooooooo + oooooooooo = ooooooooooooooooooo

Just count the circles ^

Always remember P.E.M.D.A.S (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction)

In that order, solve the problem. Multiplication and Division go left from right and so do Addition and Subtraction.

4 - 9 + ( 7 x 11 ) x 5 + 7 + 8

Parentheses

4 - 9 + (77) x 5 + 7 + 7

Multiplication

4 - 9 + 385 + 7 + 7

Addition and Subtraction

-5 + 385 + 7 + 7

380 + 7 + 7

387 + 7

394

Sorry, I just wanted to know if I could help. But if you are in a higher grade than me, then I probs can't. 

So cute! And good luck!! 

Im sure you don't need to know the grade.

I'm great, thanks. u?