All Answers 358116 Answers

Thanks chilledz3non, I didn't know that - I'll have to try and remember it!

$$y=490-\frac{2}{5}x\\\\
y= -\frac{2}{5}x+490\\\\$$

This is just a line and it is in gradient and y-intercept form. plot the y intercept which is 490. then from that point rise 200 and run 500 in the negative direction and you will have your second point. join them with a straight edge.

$$y=150-1x\\\\
y=-1x+150$$

Also  a line y intercept =150

gradient =1/1   

Yeah that's not that hard,

150mm = 0.15m

You can imagine that it is a box of which you're trying to find the volume.

The formula for a box is length*width*heigth

You know length*width = 1000

and you know height = 0.15m

So the volume you need is 1000*0.15 = 150 m^3

Reinout

All right, so here's my solution to the problem.

We apply the euclidean algorithm (I'll show you later why)

First we see how many times 19 fits into 49 and we add the remainder

Then we check how many times the remainder fits into 19 and write down a new remainder

Now we check how many times the new remainder fits into the old remainder and we keep on doing this until we have no remainder left.

49 = 19*2+11

19 = 11*1+8

11 = 8*1 + 3

8 = 3*2 + 2

3 = 2*1 + 1

2 = 1*2 + 0

Now we write the one-but-last equation into 1 = 3-2*1

Given the equation above this we can also write 1 = 3-1*(8-3*2).

This gives 1 = 3*3-8

Then again since we know 3 = 11-8*1 from the above equation

This can be rewritten as 1 = 3*(11-8*1)-8 = 3*11-4*8

Basically we keep writing the smallest number as the difference between two larger numbers following the upper equation until we have a difference of a*49-b*19 = 1

Let me put  all the steps underneath each other to make it easier to see

1 = 3-1*2

1 = 3-1*(8-3*2)

1 = 3*3-8

1 = 3*(11-8)-8

1 = 3*11-8*4

1 = 3*11-(19-11*1)*4

1 = 3*11+4*11-19*4

1 = 7*11-19*4

1 = 7*(49-19*2)-19*4

1 = 7*49-19*14-19*4

1 = 7*49-18*19

if

1 = 7*49-18*19

then

900 = 900*7*49-900*18*19

so

900 = 6300*49-16200*19

Now if we substact 19 from 6300 and 49 from 16200 the equation stays the same.

For example

900 = (6300-19)*49-(16200-49)*19

900 = 6281*49-16151*19

We want to substract 49 just 'enough' times to make it negative (so that the equation becomes an addition)

16200/49 = 330.6

So we are going to do it 331 times

then we have

900 = (6300-19*331)-(16200-49*331)*19

900 = 11*49--19*19

900 = 11*49+19*19

Now if we get back to the equation

11+19+L = 100

L = 70

Let's check that in the first equation

50*11+20*19+70 = 1000

So, she sold 11 caviar sandwiches, 19 bologna sandwiches and 70 liverwurst sandwiches

((C,B,L) = (11,19,70))

Reinout 

p.s. it is the only answer

the closest integer answers are

C = 11-19 = -8, B = 19+49 = 68 and L = 100-68+8 = 40

50*-8+20*68+40 = 1000 but C<0

and

C = 11+19 = 30, B = 19-49 = -30 and L = 100-30+30 = 100

50*30+20*-30+100 = 1000 but B<0

yes I believe it is

Energy = mass * speed of light squared

Nice answer anonymous!

Wed 28/5/14

I had fun with this one. 

I just read reinout's note. I did Bertie's money problem.  Now i am slightly more impressed.

Even though i did just use trial and error!.   

Thanks Alan,

Yes L has to be a multiple of 10 but I don't know about his part of a sandwich bit. 

--------------------------------

The reason Liverworst has to be a multiple of 10 is this.

cavier sandwiches are $5 and Bolgna sandwiches are $2.  No matter how many of the these sandwiches you buy it is going ot cost a whole number of dollars.

Altogether the sandwiches cost $100 therefore the liverworst sandwiches must cost a whole number of dollars as well. Since they are 10cents each there must be a multiple of 10 of them.

--------------------------------

Now, there are 100 sandwiches costing $100 altogether so it stands to reason there will be a lot more of the cheap liverworst ones.

I'm going to do it by emimination.

-------------------------------------------

##Say there are 90 liverworst costing $9 then there can be only 10 more sandwiches. the maximum cost of 10 sandwiches is 10*$5=$50 and that is not enough - they need to cost $100-9=$91

##Say there are 80 liverworst costing $8 then there can be only 20 more sandwiches. the maximum cost of 20 sandwiches is 20*$5=$100 umm lets look more.  

Can C+B=20 ==> 2C+2B=40 (1) and 5C+2B=92?   (2)

(2)-(1)   3C=52     52 is not a multiple of 3 so this doesn't work - C must be a whole number.

----------------------------------------

##Say there are 70 liverworst costing $7 then there can be only 30 more sandwiches. Lets look at the equations.

Can C+B=30 ==> 2C+2B=60 (1) and 5C+2B=93?   (2)

(2)-(1)   3C=33     33 is a multiple of 3 so this looks promising  33/3=11cavier sandwiches.

If this is right then there are 70liverworst, 11 cavier, and 19 bologna sandwiches.

70+11+19=100 so far so good!

5C+2B+0.1L=55+38+7=100 great.

There is the answer!   70liverworst, 11 cavier, and 19 bologna sandwiches.

I didn't check if it is the only answer but I expect that it is. It is an answer that fits the data anyway!

Have a look at this question, 

I've done such an equation before... 

-3x^2+12x-9  ?

$$\\-3x^2+12x-9 \quad | \quad : -3 \\
\boxed{x^2-4x+3=0}\\
x^2-4x \text{ } \textcolor[rgb]{1,0,0}{ + 2^2 - 2^2}+3=0\\\\
\left(x-2\right)^2-4+3=0\\\\
\left(x-2\right)^2-1=0\\\\
\left(x-2\right)^2=1\quad | \quad \pm\sqrt{} \\\\
x-2=\pm1\quad |\quad +2\\$$

$$\\x=\pm1+2\\
x_1=1+2=3 \quad\text{and}\quad x_2=-1+2=1\\
\boxed{x_1=3 \quad\text{and}\quad x_2=1}$$

$$\boxed{(-1 + i)^7 \quad? }$$

z=x+iy  $$\boxed{z=-1+i} \quad \Rightarrow x=-1, y=\textcolor[rgb]{1,0,0}{+}1$$

$$r=\sqrt{x^2+y^2}=\sqrt{(-1)^2+1^2}=\sqrt{2}$$

$$\phi = sign(y)*\cos^{-1}{\left(\frac{x}{r}\right)}
=\textcolor[rgb]{1,0,0}{+}\cos^{-1}(\frac{-1}{\sqrt{2}})
=\cos^{-1}(-\frac{\sqrt{2}}{2})=\frac{3}{4}\pi$$

$$\boxed{z=r*e^{i\phi}}$$

$$\\z=\sqrt2*e^{i*\frac{3}{4}\pi} \\
z^7=\left(\sqrt2*e^{i*\frac{3}{4}\pi}\right)^7 \\
z^7=\left(\sqrt2\right)^7*\left(e^{i*\frac{3}{4}\pi}\right)^7 \\
z^7=\left(\sqrt2\right)^7*e^{i*\frac{7*3}{4}\pi} \\
z^7=\left(\sqrt2\right)^7*e^{i*\frac{21}{4}\pi}
= \left(\sqrt2\right)^7*e^{i\left(4\pi+\pi+\frac{1}{4}\pi\textcolor[rgb]{1,0,0}{-4\pi}}\right)\\$$

$$z^7=\left(\sqrt2\right)^7*e^{i\left(\pi+\frac{1}{4}\pi}\right)\\$$

$$\boxed{e^{i\phi}=\cos{\phi}+i\sin{\phi}}$$

$$\\z^7=\left(\sqrt2\right)^7 \left[ \cos{(\pi+\frac{\pi}{4} )}+i*sin{(\pi+\frac{\pi}{4})} \right]\\
z^7=\left(\sqrt2\right)^7
\left[
\underbrace{\cos{\pi}}_{-1}\cos{\frac{\pi}{4}} - \underbrace{\sin{\pi}}_0\sin{\frac{\pi}{4}}
+i\left( \underbrace{\sin{\pi}}_0\cos{\frac{\pi}{4}}+\underbrace{\cos{\pi}}_{-1}\sin{\frac{\pi}{4}} \right)
\right]\\
z^7=\left(\sqrt2\right)^7 \left( -\cos{\frac{\pi}{4}}-i\sin{\frac{\pi}{4}}\right)$$

$$\sin{\frac{\pi}{4}}=\cos{\frac{\pi}{4}}=\frac{\sqrt{2} }{2}\quad\\
z^7=\left(\sqrt2\right)^7 \left( -\frac{\sqrt{2}}{2}-i{\frac{\sqrt{2}}{2}\right)\\
z^7=\left(\sqrt2\right)^7\frac{\sqrt{2}}{2}\left( -1-i\right)\\
z^7=\frac{\left(\sqrt{2}\right)^8}{2}\left(-1-i\right)$$

$$\\z^7=\frac{2^4}{2}\left(-1-i\right)\\\\
z^7=2^3\left(-1-i\right)\\
z^7=8\left(-1-i\right)\\
z^7=-8-8i\\$$

$$\boxed{(-1+i)^7=-8-8i}$$

Note that L must be a multiple of 10 to avoid having a fraction of a sandwich!  (Hint: try L = 70).

let us do it like this ;

as 100 is the  whole percen so 

75/100 x 420

cut  100 and 420 by 100 being 5 and 420 being 21 .

75/5 x 21

cut 75 by 5 with 75 being 15 

so now u have

15 x 21

=315

so 75 percent of 420 is 315 ! 

Calculus is a huge and very powerful branch of mathematics.

To ask how do i solve calculus is to ask how do i solve algebra.

The question makes no sense!

owen made 100 sandwiches which she sold for exactly $100.

She sold caviar sandwiches for $5.00 each,

the bologna sandwiches for $2.00, and the liverwurst sandwiches for 10 cents.

How many of each type of sandwich did she make?

5C+2B+0.10L=100 => 50C+20B+L=1000 (1)

C+B+L=100 (2)

These are simultaneous equations and nomally they could not be done because there are 3 unknows so you need 3 equations.  This is a little different because C, B and L must be whole numbers.

This means that this is a LINEAR DIOPHANTINE EQUATION.

(1)-(2) = 49C+19B=900

 Maybe you can solve this with trial and error???

To solve it properly I will have to do some serious homework.  (Which given enough time I will do)

$${\left({\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{i}\right)}^{{\mathtt{7}}} = {\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{8}}{i}$$

$${expand}{\left({\left({\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{i}\right)}^{{\mathtt{7}}}\right)} = {{i}}^{{\mathtt{7}}}{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{6}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{5}}}{\mathtt{\,-\,}}{\mathtt{35}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{35}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{21}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}$$

$$\sqrt[3]{-27}=\sqrt[3]{27}\sqrt[3]{-1}=3\sqrt[3]{-1}$$

$$\sqrt[3]{-1}=\sqrt[3]{e^{\imath \pi}}=\sqrt[3]{e^{\imath (\pi +2k \pi)}}~~k\in \mathbb{Z}$$

and we pick values corresponding to k=0,1,2

$$\sqrt[3]{-1}=e^{\imath \pi/3}$$

$$\sqrt[3]{-1}=e^{\frac{\imath(\pi + 2\pi)}{3}}=e^{\imath\pi}=-1$$

$$\sqrt[3]{-1}=e^{\frac{\imath(\pi + 4\pi)}{3}}=e^{\imath 5\pi/3}=e^{-\imath \pi/3}$$

so

$$\sqrt[3]{-27}=-3, ~3e^{\imath \pi/3}, ~3e^{-\imath \pi/3}$$

Oh I see it wants the answers in rectangular form

To do this, convert the complex exponential into rectangular form as follows

$$e^{\imath x}=\cos(x)+\imath \sin(x)$$

$$-3 = -3 \\$$

$$3e^{\imath \pi/3} = 3(1/2 + \imath \sqrt{3}/2)=3/2+\imath 3\sqrt{3}/3$$

$$3e^{-\imath \pi/3} = 3(1/2 - \imath \sqrt{3}/2)=3/2-\imath 3\sqrt{3}/3$$

$$\boxed{2^{\frac{x}{2}} * 4^{\frac{x}{6}}* \left[(\frac{1}{8})^{\frac{1}{x}}\right]^{\frac{1}{6}}=2^2*2^{\frac{1}{3}} }\\\\
\Rightarrow 2^{\frac{x}{2}} *2^{2*\frac{x}{6}}*2^{-3*\frac{1}{x}*\frac{1}{6}}=2^{2+\frac{1}{3}}\\\\
\Rightarrow 2^{\frac{x}{2}} *2^{\frac{x}{3}}*2^{\left(-\frac{1}{2x}\right)}=2^{\frac{7}{3}}\\\\
\Rightarrow 2^{
\left(\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right)
}=2^{\frac{7}{3}} \quad | \quad ln\\\\
\Rightarrow {
\left(\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right)*\ln(2)
}=\frac{7}{3}*\ln(2)\\\\
\Rightarrow {
\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right
}=\frac{7}{3}\\\\$$

$$\\\Rightarrow {
\frac{3x}{3x}*\frac{x}{2}+\frac{2x}{2x}*\frac{x}{3}
-\frac{3}{3}*\frac{1}{2x}\right
}=\frac{7}{3}\\\\
\Rightarrow \frac{3x^2+2x^2-3}{6x}=\frac{7}{3} \quad | \quad *6x\\\\
\Rightarrow 5x^2-3=6x*\frac{7}{3}\\
\Rightarrow 5x^2-3=14x\\$$

$$\\\Rightarrow 5x^2-14x-3=0\\\\
\Rightarrow x^2-\frac{14}{5}x-\frac{3}{5}=0\\\\
\Rightarrow x_{1,2}=\frac{14}{2*5}\pm\sqrt{\frac{14*14}{(2*5)*(2*5)}+\frac{3}{5}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{14*14}{(10)*(10)}+\frac{20}{20}*\frac{3}{5}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{196}{100}+\frac{60}{100}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{196+60}{100}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{256}{100}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\frac{16}{10}\\\\
\textcolor[rgb]{1,0,0}{x_1=}\frac{14}{10}+\frac{16}{10}=\frac{30}{10}=\textcolor[rgb]{1,0,0}{3}\\\\
\textcolor[rgb]{1,0,0}{x_2=}\frac{14}{10}-\frac{16}{10}=-\frac{2}{10}=\textcolor[rgb]{1,0,0}{-\frac{1}{5}}\\\\$$

On this site's scientific calculator at least, just press "Enter" twice, and the second time the answer will be in scientific notation.

The formula for circumference is:  $${\mathtt{C}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{d}}$$ where d is the diameter of the circle.

Therefore, plug in 4 for d, and the circumference would equal  $${\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}} = {\mathtt{12.566\: \!370\: \!614\: \!359\: \!173}}$$. You can write this either as 4 pi or round the decimal.

The formula for area of a circle is: $${\mathtt{A}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{2}}}$$ where r is the radius of the circle. Since radius is half of the diameter, the radius of this circle = 2

And 2² is still 4. So the area would be 4 pi as well, about  12.57.

The difference between the two values for circumference and area of this circle is their units. Area is in units squared.

Since 75% is the same as 0.75 (< the decimal form), simply multiply 420 by 0.75.

$${\mathtt{420}}{\mathtt{\,\times\,}}{\mathtt{0.75}} = {\mathtt{315}}$$

The Pythagorean Theorem is named for the Greek mathematician Pythagoras.

It uses the equation   $${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{c}}}^{{\mathtt{2}}}$$   to solve for lengths of a right triangle's sides, given two sides and one unknown.

A and b would be the lengths of the triangle's legs (two shorter sides) and c would be the length of the hypotenuse (longest side, diagonal to legs).

You can read more about it here:

http://www.purplemath.com/modules/pythagthm.htm

step1:  subtract 1.6x from both sides

step2: subtract 5.4 from both sides.

step 3: decide what you need to divide (both sides) by to get rid of the number in front of the x

Try it yourself.

Start by multiplying both sides by 4