Straight

I'm problably too late, but I am still answering,

Answer: B because by Jane (\(10*3\)) is 30 and then subtract the $20 and this is 10,

and by Chris is the same, but (\(10*5\)) is 50 and then subtract the $40 and this is also 10.

By A: Jane (\(8*3\)) is 24 and then subtract the $20 and this is 4,

and by Chris the same, but (\(8*5\)) is 40 and then subtract the $40 and this is 0 (difference: 4).

By C: Jane (\(12*3\)) is 36 and then subtract the $20 and this is 16,

and by Chris the same, but (\(12*5\)) is 60 and then subtract the $40 and this is 20 (difference: 40).

By D: Jane (\(20*3\)) is 60 and then subtract the $20 and this is 40,

and by Chris the same, but (\(20*5\)) is 100 and then subtract the $40 and this is 60 (difference: 60).

Straight

There is no solution, if there are no values (except in the headline, but not in all).

Straight

Hello,

from one to a hundred, 10 numbers have the number 6 as the ones digit, whereas a zero is enough for 66.

Now the tens digit could also be a 6, i.e. a total of 10, which means that there are 19 numbers,

from one to one hundred, that have at least one 6 (6,16,26,36,46,56,60,61,62,63,64,65,66,67,68,69,76,86,96).

From 100 to 200, only 10 numbers can have a 6 in the ones place and 10 in the tens place and the hundreds place 0,

because it only starts with a one. Finally, there are 19 numbers that have at least one 6

(106,116,126,136,146,156,160,161,162,163,164,165,166,167,168,169,176,186,196).

From 200 to 300 there are also only 19 numbers, from 300 to 400 also only 19 numbers,

from 400 to 500 also only 19 numbers, from 500 to 600 also only 19 numbers,

but from 600 to 700 all numbers that total 100 have a 6,

then from 700 to 800 also only 19 numbers. Finally, this gives (\(19+19+19+19+19+19+100+19=233 \)) numbers.

Straight

Take the largest natural numbers from the set (2)

and multiply them by the smallest rational number (1), which gives -448. 

(\(7*8*-8=-448\)).

Straight

\(40:2\) is \(20\) ,

the half of 40 is 20.

Straight

There is a hard methode to get \(\sqrt{53}\) , I have left it out

The maximum value for \(wxyz\), where it applies \(w+x+y+z=8\) ,

is 16 (\(2+2+2+2=8\)) and (\(2*2*2*2=16\)).

Let's look, how I did this:

If \(w\) (or \(x,y,z\)) is 8 and the others are only 0, this wouldn't be possible because 8*0*0*0 = 8,

If \(w\) (or \(x,y,z\)) is 7 can't stand 0,7 or like that because that makes the number smaller and isn't a real number.

So 7*1*0*0 is still 0 because if at least one factor a zero, then the product is ALWAYS 0,

so we have to distribute the factors with integers that are also not 0. After this rule, we're starting now with 4:

(by 5 and 6 is the product 0, so it's not working there)

4*1*1*2 = 8 ( - too small),

let us look by 3:

3*2*1*2 = 12 ( - too small for 16) or other possibilities would be the product (\(n<13\)).

By 2 it works great:

(\(2^4\)) or (\(2*2*2*2 \)) eqauls 16. There is no possibility that equals 16, except one.

By 1 isn't it possible (reason: The product of \(1*x*y*z\) is maximum \(n<13\) ).

So if you don't know what's the maximum of \(w*x*y*z\) is 16 (\(2*2*2*2\) ).

Straight

Hello,

(\(x+y=5\)) and (\(xy=-7\)),

(\(x=5-y\)) and (\(xy=-7\)),

\((5-y)y=-7\)),

so y is now:

(\(y=\frac{5+\sqrt{53}}{2}\)) and (\(y=\frac{5-\sqrt{53}}{2}\)),

(\(x=5-\frac{5+\sqrt{53}}{2}\)) and (\(y=\frac{5-\sqrt{53}}{2}\)),

(\(x=5-\frac{5+\sqrt{53}}{2}\)) and (\(x=5-\frac{5-\sqrt{53}}{2}\)),

so x is now:

(\(x = \frac{5-\sqrt{53}}{2}\)) and (\(x = 5-\frac{5-\sqrt{53}}{2}\)),

(\(x = \frac{5-\sqrt{53}}{2}\)) and (\(x = \frac{5+\sqrt{53}}{2}\)),

So there are 2 possibilities, namely:

\(x,y=(\frac{5-\sqrt{53}}{2}, \frac{5+\sqrt{53}}{2})\) and the second possibility, namely:

\(x,y=(\frac{5+\sqrt{53}​​​​​​​​}{2}, \frac{5-\sqrt{53}}{2})\) 

Straight

The average is:

(\(89+69+56+23+58+69=364\)),

(\(\frac{364}{6}=\frac{182}{3}=60\frac{2}{3}=60.\overline6\))

The average is (\(60.\overline6\)).

Straight