All Answers 356020 Answers

135 choco. candies

Fifty ways to choose a state's senators   then 49 ways to fill the third seat

50 * 49 = ...............

System of equations

2/5 c + 1/2 s = 138    <==== solve this for   s= 2 (138 -2/5 c)  

2/3 c + 5/7 s = 210    < ==== sub in the red equation for s

  2/3 c + 5/7 (2(138-2/5 c) )  = 210        now solve for 'c'

0.334 is greater than 0.33 or 0.330

How much is the basketball?

Thanks

x = amount at first 

{x  -  (1/4x-25) } *2/5  -17 - 18   = 65

(3/4x +25) 2/5  =100

x = 300 dollars

8 x 8 = 64 possible rolls

2 ways to roll 15          8 7      7 8

2/64 =1/32

Using different units:

m / v = density

5.98* 10^24 kg  /  4/3 pi (6.38* 10^8)^3   = .0054973  kg/m³   =5.49 *10^-3  kg/m³

5th term + 23d   = 32nd term

9   + 23d = 88

  d = 79/23

5th term + 18d = 23rd term

9 + 18 * ( 79/23) = ...........

circumference = pi * d

240 = pi * d

d = 240/pi    

    radius = 1/2 d     =   120/pi

Area = pi r^2

            = pi ( 120/pi)^2   =  14400/pi ft²

Assuming all the animals are classed as individuals it really makes no difference what species that they are.

There are 15 animals.  12 of which can be on a committee with any other one.

So I have split it up into 6 from the non-choosy ones + (a fussy one*5from the non-choosy ones) *3

12C6 + 3*12C5 =  3300  committies

100pi=314.159265359

radius would equal 50

hint: two triangles are congruent if they have 2 sides and the included angle the same size.  SAS test

-10m + 8 <= 10m - 2

\(-10m + 8 \le 10m - 2\\ -20m + 8 \le - 2\\ \qquad -20m \le -10\\ \qquad \qquad m\ge0.5\)

The mass of tithe earth is 5.98x10^24kg, and its radius is 6.38x10^8m. Compute the density of the earth, using powers-of-ten notation and the correct number of significant figures.

Hello Guest!

The radius of our earth is \(6393km \approx 6.39 \times10 ^ 6m\).

\(Density=\dfrac{mass}{volume}\)

\(\rho=\dfrac{m}{V}=\dfrac{5.98\cdot 10^{24}\cdot kg}{\frac{4}{3}\pi\cdot 6.39^3\cdot (10^6)^3\cdot m^3}\times\dfrac{m^3}{10^6\cdot cm^3}\times\dfrac{10^3\cdot g}{kg}\\ =\dfrac{3\cdot 5.98}{4\cdot 6.39^3\cdot \pi}\times\dfrac{10^{24}\cdot 10^3}{10^{18}\cdot 10^6}\times\dfrac{g}{cm^3}\color{blue}\approx5.47g/cm^3\)

  !

Thanks Alan,

That works  :)     

Guest's answer is exactly 3 times more than yours.  That suggests more than just a coincidence....  

Here is my best guess

Say they all go consecutive (no pairs)

5! ways to order them then each of those can go to pad 1 or pad 2.

So that is 5! 8 2^5 = 120*32 = 3840 ways

Now look at one pair.  

There are 5C2=10 ways to choose the pair  (ty these together with a rope so now they only count as one.

There are 4! ways to line them up now.

the three singles will each have a choice of pad that is 2^3 ways

The double obviously uses both pads but can be in either order on those pads so that is 2 ways

10 * 4! * 2^3 * 2 = 10 * 24 * 16 = 3840 ways

Now look at 2 pairs

I think that there are  5C2 * 3 /2 ways to chose these = 15 ways

there are 3! ways to line them up

there are 2^3 ways to chose the pads.

15 * 6 * 8 = 720 ways

3840+ 3840 + 720 = 8400 ways 

Using Mathcad to do the heavy lifting I get the following:

I suspect there is a more elegant way, but I can;t think of one!

Mmm

if the small one is 10  then the other can be 11,12 or 13   That is 3 possibilities

---

from 11 to 95 inclusive there is 95-11+1=85 numbers each with 3 possible matches

---

if the small one is 96  then the other can be 97, 98, 99   That is 3 possibilities

if the small one is 97  then the other can be 98,99   That is 2 possibilities

if the small one is 98  then the other can be 99   That is 1 possibilities

So I have  87*3+2+1 = 264 favourable pairs

Now there are 90 numbers to start with.  So 90C2= 4005  pairs are in the sample space

so the probability of the desired event happening is     264/4005 = 88/1335

I have not checked for silly errors.

In triangle ABC, find sin A if b = 2*sqrt(10), c = 11, and a = 9.

Hello Guest!

\(a^2=b^2+c^2-2bc\ cosA\\ {\color{blue}cosA}=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{4\cdot 10+11^2-9^2}{2\cdot 11\cdot 2\cdot \sqrt{10}}=\dfrac{80}{44\cdot \sqrt 10}=\color{blue}\dfrac{20}{11\cdot \sqrt{10} }\)

\(sinA=\sqrt{1-cos^2A}=\sqrt{1-\dfrac{400}{121\cdot 10}}=\sqrt{\dfrac{81}{121}}\\ \color{blue}sinA= \dfrac{9}{11}=0.\overline{18}\)

  !

55*1=55 mod 220

55*2=110 mod220

55*3=165mod220

55*4=0 mod220

so

55*(4k+3)=165(mod220)

n=4k+3

the biggest three digit n is 999

I have no idea how to do this.  So if a mathematician wants to explain it that would be really good.

Of course, if the coefficients did not have to be integers then a could be as close to 0 as you please.

I did construct a graph of degree 7 but it is not actually helpful

Here is a link:

sqrt[17² - (x/2)²] = sqrt[27² - (1.5x)²]

x = 2√55           BD = 2x

AF = sqrt[17² - (x/2)²]

[ABD] = 1/2(AF * BD)