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5N+10D=235

Do you mean shift caret on your computer keyboard 

The caret (or hat) is the symbol for exponent.

for instance if you input 5^3 then this is what you mean

$$5^3 = 5*5*5 = 125$$

Thanks Civonamuk ,

Here is another way to tackle this question. :)

6% is   57c

1% is   57/6 cents

100% is    57/6*100cents

$${\frac{{\mathtt{57}}}{{\mathtt{6}}}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{950}}$$   cents

original pay was   $9.50 per hour

Thanks you Alan.

"The cross-sectional area is the area perpendicular to the direction of current flow, through which the current is flowing.  If the current were flowing axially through a solid cylinder of radius r, then the cross-sectional area would, indeed, be pi*r², and it would be constant all the way along the cylinder.  However, here the current is flowing radially outward , so the cross-sectional area gets bigger as you go further out from the centre."

I do not understand the difference between these 2 highlighted terms.  

(It is probably because I have absolutely no understanding of electricity at all)

0 ways...

To see why.......the sum of three odd numbers would have to equal an even, and this is impossible. Ths sum of three odd numbers is always odd.

F = G*m1*m2/r², where F is force, G is gravitational constant(6.673×10⁻¹¹ N·(m/kg)²), m's are masses and r is distance between masses.

$${\mathtt{1.27}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{8}}} = {\frac{{\mathtt{6.673}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{11}}\right)}{\mathtt{\,\times\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{m}}}{{{\mathtt{5}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{m}} = {\frac{{\mathtt{105\,151\,356\,211\,599}}}{{\mathtt{1\,105\,000\,000\,000}}}} \Rightarrow {\mathtt{m}} = {\mathtt{95.159\: \!598\: \!381\: \!537\: \!556\: \!6}}$$

The other mass is approximately 95 kg

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 y=-3x^2+12x+1      there is no "solution" to this.....it's the equation for a parabola.....see it here.....

c. The distance of each ball from the Earth's center of mass is almost identical.

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a.  0 J

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$$\\
\frac{\cos(x)\sin^2(x)}{1-\cos(x)}=\cos(x)+\cos^2(x) \\\\
\text{Divide by cos(x)}\\
\frac{\sin^2(x)}{1-\cos(x)}=1+\cos(x) \\\\
\text{Multiply by 1-cos(x)}\\
\sin^2(x)=(1-\cos(x))(1+\cos(x))\\\\
\text{The right-hand side is the difference of two squares so}\\
\sin^2(x)=1-\cos^2(x)\\\\
\text{so:}\\
\sin^2(x)+\cos^2(x)=1\\
\text{which is correct!}$$

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(x-2)(x+4)(x-5)^2<0

The answer will come from one - or more - of these intervals  (-∞, -4), (-4, 2), (2,∞)

Notice that any x values on   (-∞, -4). (2, ∞)  make the equality false

And any x values on (-4, 2)  make it true

Use the sine rule:  sin(A)/a = sin(B)/b etc.

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The inequality holds when -4 < x < 2

Look at the graph below and think carefully about what happens in each region:

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You are correct Melody and xvxvxv.

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Many more people have been killed and tortured in the name of Christianity then have been killed in the name of Islam.

If you want to belong to a religion that has not been responsible for terrible atrocities then DO  NOT be a Christian or a Muslim!

Buddhism seems like a really nice religion.  Maybe become a Buddhist.   

I am not religious at all.  Therefore I am not biased.

5000(1.07)^2

They are ususally just brackets.

But it would depend on the context  

subtract 6 from both sides

Not quite MG  powers must be done before multiply

$${\mathtt{6}}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{7}}} = {\mathtt{1\,679\,616}}$$

you  could have given some working Brittany   

so could you anon :)

Don't fret I will answer your question :)

No te preocupes voy a contestar su pregunta :)

counting techniques: A math test has 15 questions which must answer 10. from how many different ways you can select questions?

15C10

input       nCr(15,10)

output      $${\left({\frac{{\mathtt{15}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)} = {\mathtt{3\,003}}$$

You cannot divide by 0 so it is undefined  :)

3/x+1/2=1

subtract a half from both sides

then turn both sides upside down

then what ??