In order to solve this problem, set up the following equation:

$\displaystyle \frac{1.20}{x}=\frac{150\%}{100\%}$

Cross multiply:

$\displaystyle 150x = 120$

Divide:

$\displaystyle \frac{150x}{150} = \frac{120}{150} = 0.80$

The original cost of the of each candy bar is $\displaystyle \$0.80$

$\displaystyle \frac{12x^3y^2p}{4ab}\div\frac{3x^3y}{2ab}$

To divide monomial quotients, simply invert the divisor and multiply:

$\displaystyle \frac{12x^3y^2p}{4ab} *\frac{2ab}{3x^3y} = \frac{24x^3y^2pab}{12abx^3y}$

Then, reduce:

$\displaystyle \frac{24yp}{12} = 2yp$

$\displaystyle \frac{10m^2np}{3xy}\div\frac{3p^2}{2xym}$

To find your answer, you have to invert the divisor and multiply across:

$\displaystyle \frac{10m^2np}{3xy} * \frac{2xym}{3p^2} = \frac{20m^3npxy}{9p^2xy}$

Then, reduce:

$\displaystyle \frac{20m^3n}{9p}$

To solve you must multiply $\displaystyle 2x$ by both terms in $\displaystyle (4x+3)$

$\displaystyle 2x\ast4x=8x^{2}$

$\displaystyle 2x\ast3=6x$

$\displaystyle 8x^{2}+6x$

Multiply $\displaystyle 4x$ by both terms in $\displaystyle (5x+7)$

$\displaystyle 4x\cdot5x=20x^{2}$

$\displaystyle 4x\cdot7=28x$

$\displaystyle 20x^{2}+28x$

When multiplying a polynomial by a monomial, each term in the polynomial gets multiplied by the monomial. Calculate each term one at a time, then add the results to get the final answer. In this case, we start by multiplying $\displaystyle 7x^{2}\cdot3x$. $\displaystyle 7\cdot3=21$ and $\displaystyle x^{2}\cdot x=x^{3}$, thus we get $\displaystyle 21x^{3}$. For the second term of the polynomial, we multiply $\displaystyle -12\cdot3=-36$ and $\displaystyle x\cdot x= x^{2}$, resulting in $\displaystyle -36x^{2}$. Finally, we multiply $\displaystyle 4\cdot3 = 12$ and $\displaystyle 1\cdot x = x$, resulting in $\displaystyle 12x$. Adding the three terms that we just found, we come to the answer of $\displaystyle 21x^{3}-36x^{2}+12x$.

$\displaystyle \frac{4x^2p^3}{3mn^2} * \frac{2mn}{12xy}$

To multiply monomial quotients, treat them as you would any other fraction. Combine like terms wherever possible:

$\displaystyle \frac{8x^2p^3mn}{36mn^2xy}$

Then, you need to reduce:

$\displaystyle \frac{2xp^3}{9ny}$

$\displaystyle \frac{2x^3y^4}{10z^3} * \frac{4z^2p}{10xy}$

To simplify, first multiply across:

$\displaystyle \frac{8x^3y^4z^2p}{100z^3xy}$

Then, reduce:

$\displaystyle \frac{2x^2y^3p}{25z}$

This is a direct variation problem of the form y = kx²  The first set of data 3.6 g and $64.80 is used to calculate the proportionality constant, k.  So 64.80 = k(3.6)² and solving the equation gives k = 5.

Now we move to the new data, 7.5 g and we get y = 5(7.5)² to yield an answer of $218.25.

$135.00 is the answer obtained if using proportions.  This is an error because it does not take into consideration the squared elements of the problem.

The question asks for an equation that can relate $\displaystyle d$ and $\displaystyle v$ to each other, based on the information given. We are told that half the volume + $\displaystyle 6cm$ determines the total diameter.

This gives us:

$\displaystyle d=\frac{1}{2}v+6$