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I'm impressed Chris :)

check

$${\mathtt{0.5}}{\mathtt{\,\times\,}}\left({\mathtt{26}}{\mathtt{\,\small\textbf+\,}}{\mathtt{40}}{\mathtt{\,\small\textbf+\,}}{\mathtt{64.78}}\right) = {\frac{{\mathtt{6\,539}}}{{\mathtt{100}}}} = {\mathtt{65.39}}$$

$${\sqrt{{\mathtt{65.39}}{\mathtt{\,\times\,}}\left({\mathtt{65.39}}{\mathtt{\,-\,}}{\mathtt{26}}\right){\mathtt{\,\times\,}}\left({\mathtt{65.39}}{\mathtt{\,-\,}}{\mathtt{40}}\right){\mathtt{\,\times\,}}\left({\mathtt{65.39}}{\mathtt{\,-\,}}{\mathtt{64.78}}\right)}} = {\mathtt{199.730\: \!747\: \!341\: \!489\: \!712\: \!9}}$$

yes that is correct

which is an equation of the line that is parallel to y = 2x - 8 and passes through the point (0,-3) ?

a) y = 2x + 3    b) y = 2x - 3    c) y = -1/2x + 3    d) y = -1/2x - 3

Well if it is parallel then it will have the same gradient   m=2

so it has to be a or b

sub in (0,-3) and see if it makes the equation true.

a) -3=2*0+3    not true

b) -3=2*0-3     true

so the answer is b

Use the calc on the home page. do not forget the * sign

Thanks Chris :)

The equation  y - 2x  =  4  can be written as  y  =  2x + 4

This line has a slope of 2.

Any line perpendicular to this line has a negative reciprocal slope; the negative reciprocal of 2 is -1/2.

Which answer has a slope of -1/2?

Use the formula:  A  =  P(1 + r)^t

where A = final amount = 283400     P = initial amount = 200300     r  =  rate    t  = number of years = 10

283400  =  200300(1 + r)¹⁰

Divide by 200400    --->   1.414877683  =  (1 + r)¹⁰

Find the 10th root   --->   1.0353  =  1 + r

Subtract 1              --->   r  =  0.0353  or  3.53%

think of negative numbers as cold numbers and positive numbers as hot numbers.

if you take away cold from a room then the room will get hotter.  so     - - 3 = +3

- 2 - - 3 = -2+3 =1

to get the last bit, start at -2 on the number line and go 3 places in the positive direction.

You end up at 1.

3x+4y+2x+5y

=3x+2x    +4y+5y

=5x  +  9y

10 x 10 is written as  10*10  and it equals 100

10²  is writen as 10^2   it is also equal to 10*10 and equals 100

Inverse trig is use to find angles.

Inverse trig is the same as arc trig  so on this calc you will use   asin

$$sin^{-1}(0.42)=asin(0.42)$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.42}}\right)} = {\mathtt{24.834\: \!587\: \!489\: \!702^{\circ}}}$$

English translation

Is this the new 10+9 ?

I would think so. :)

There is no fomula, you just add up the sides.

Do you meant a regular nonogon, where all the sides are the same and all the angles are the same?

Hi Alan,

I have not seen probability dealt with like this before.

I can understand part A and it sort of makes sense to me I think.

I am trying to work out the mathematical logic of what you have done it part b.

I get a glimmer of understanding but then it goes again.  Can you help me to understand this?

Thank you.

it is 100 not 1

If we do this correctly....we will arrive at a familiar formula....let's see...

ax^2+bx+c=0   subtract c from both sides

ax^2 + bx  = -c       divide through by a

x^2 +(b/a)x = -c/a     complete the square on x

x^2 + (b/a)x + [b^2] / [4a^2]  = -c/a + [b^2] / [4a^2]     factor the right side....get a common denominator on the left

[x + (b/2a)]^2  = [b^2 - 4ac] / [4a^2]   take the square roots of both sides

x + (b/2a)  = ±√[b^2 - 4ac]/[2a]    subtract (b/2a) from both sides

x = (-b±√[b^2 - 4ac]) / [2a]

Voila!!!....we have the Quadratic Formula..... !!!!

$${\left({{\mathtt{x}}}^{-{\mathtt{3}}}\right)}^{{\mathtt{4}}}$$

$${\left({\frac{{\mathtt{1}}}{{{\mathtt{x}}}^{{\mathtt{3}}}}}\right)}^{{\mathtt{4}}}$$

$${\frac{{{\mathtt{1}}}^{{\mathtt{4}}}}{{\left({{\mathtt{x}}}^{{\mathtt{3}}}\right)}^{{\mathtt{4}}}}}$$

$${\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}{{\left({\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}^{{\mathtt{4}}}}}$$

$${\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}{\left({{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{4}}}\right)}}$$

$${\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}{\left(\left({\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)\right)}}$$

$${\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}{\left({\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}}$$

$$\left({\frac{{\mathtt{1}}}{{{\mathtt{x}}}^{{\mathtt{12}}}}}\right)$$

It's 6x when it is an expopnent

$${\mathtt{\,-\,}}{\sqrt{-{\mathtt{308\,025}}}}$$

$${\mathtt{\,-\,}}{\sqrt{{\mathtt{37}}{\mathtt{\,\times\,}}{\mathtt{37}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}-{\mathtt{1}}}}$$

$${\mathtt{\,-\,}}\left({\sqrt{{\mathtt{37}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{37}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\sqrt{-{\mathtt{1}}}}\right)$$

$${\mathtt{\,-\,}}\left({\mathtt{37}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{-{\mathtt{1}}}}\right)$$

$${\mathtt{\,-\,}}\left({\mathtt{555}}{\mathtt{\,\times\,}}{\sqrt{-{\mathtt{1}}}}\right)$$

$$\left(-{\mathtt{555}}{i}\right)$$

$$-{\mathtt{555}}{i}$$

We'll have to use Heron's Formula to solve this.

Let  a = 26, b= 40 and x be the unknown side

s = the semi-perimeter  =  [x + 26 + 40]/2 = [x + 66]/2

So we have

200 = √[s(s -a)(s-b)(s-x)]

200^2  = [x + 66]/2 * [(x + 66)/2 - 26] * [(x + 66)/2 - 40]* [(x + 66)/2 - x] 

200^2 = [x + 66]/2 * [ x + 14]/2 * [x - 14]/2 * [66 - x]/2

200^2 = [x + 66] * [x + 14] * [x - 14] * [66- x] * (1/16)

40000 = [x + 66] * [x + 14] * [x - 14] * [66- x] * (1/16)

640000= [x + 66] * [x + 14] * [x - 14] * [66- x] 

We could brute force this.....but....I'm going to let WolframAlpha do the heavy lifting...

It returns two solutions

x = 2√89 ≈ 18.87   and  2√1049 ≈ 64.78

Both satisfy the Triangle Inequality.....so.....looking at the Law of Cosines

18.87^2  = 40^2 + 26^2 - 2(40)(26)cosΘ   

And the angle between the sides of 26 and 40 could be 22.6°

And using the Law of Sines, we have

sinΘ/40 = sin 22.6/18.87

sinΘ = 40sin22.6/18.87 =  54.55°

And the remaining angle is 180 - 22.6 - 54.55 = 102.85

But this is impossible because it would mean that the greatest angle is opposite the intermediate side

And using the Law of Cosines again, we have... 

64.78^2 = 40^2 + 26^2 - 2(40)(26)cosΘ 

So, the angle between the sides of 26 and 40 could be 157.38°

And using the Law of Sines again, we have

sinΘ/40 = sin 157.38/64.78 = 13.73°

And the remaining angle is 180 - 157.38 - 13.73 = 8.89°

So, the solution is

Side 64.78,  opposite angle = 157.38°

Side 40, oppsite angle 13.73°

Side 26, opposite angle 8.89°

So....the remaining unknown side is 64.78

Here's the approximate triangle

SEE MELODY'S ANSWER BELOW...THE SECOND SOLUTION IS POSSIBLE  !!!

The 3 at the front accounts for the fact that the girl can be the first or the second or the 3rd child.

If the three was not there you would have the prob that the girl was in a specific position. 1st or second or 3rd.

If found by formula it would be     3C1 one girl out of three    (or 3C2  two boys out of 3)  they are the same.

$$3C1=\frac{3!}{1!(3-1)!}=\frac{3!}{2!}=3$$

$${{\mathtt{5}}}^{-{\mathtt{2}}}$$

$${\frac{{\mathtt{1}}}{{{\mathtt{5}}}^{{\mathtt{2}}}}}$$

$${\frac{{\mathtt{1}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}}$$

$${\frac{{\mathtt{1}}}{{\mathtt{25}}}}$$

boy boy girl   0.51*0.51*0.49

boy girl boy   0.51*0.49*0.51

girl boy boy  0.49*0.51*0.51

So probability is

$${\mathtt{p}} = {\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{0.51}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{0.49}} \Rightarrow {\mathtt{p}} = {\mathtt{0.382\: \!347}}$$

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yarrrrrrrak