Will85237

\(\frac{(12a^4b^5)^3}{(12a^6b^8)^8}\)

\(=\frac{1728a^{12}b^{15}}{1728a^{48}b^{64}}\)

\(=\frac{1}{a^{36}b^{49}}\)

\(\frac{14}{20}\times100\)

\(=70\%\)

You need another equation in order to solve with matrices, unless you just want to express x in terms of y or vice versa.

Yes, fractions are the same as dividing.

Any number you want.

Just take away 70 one cent coins and add 2 ten cents and 2 twenty-five cents then.

\((c - 4d)^3 \)

\(=(c)^3+3(c)^2(-4d)+3(c)(-4d)^2+(-4d)^3\)

\(=c^3-12c^2d+48cd^2-64d^3\)

\(3 sec^2 (x) - 2\sqrt{3} tan (x) - 6 = 0\)

Using the identity sec^2(x) = 1+tan^2(x)

\(3(1+tan^2(x)) - 2\sqrt{3} tan (x) - 6 = 0\)

\(3tan^2(x) - 2\sqrt{3} tan (x) - 3 = 0\)

\(3tan^2(x) - 2\sqrt{3} tan (x) - 3 = 0\)

\((3tan^2(x)+\sqrt{3})(tan^2(x)-\sqrt{3})=0\)

\(tan(x)=\sqrt{-\sqrt{3}/3}\)

\(no \space solution\)

\(tan(x)=\sqrt[4]{3}\)

\(x=52.8°\space or \space 233°\)

If you think about how this would expand, x^99 can only be created when 99 xs multiply, times a constant.

Because there are 100 different constants (-1 to -100), the coefficent will be the sum of an arithmetic sequence.

\(-1\times x^{99}-2\times x^{99}-3\times x^{99}-...-100\times x^{99}\)

\(S_{100}=\frac{100}{2}(-1-100)\)

\(S_{100}=-5050\)

Coefficient will be -5050.

Let w = number of 1 cent coins, x = number of 5 cent coins, y = number of 10 cent coins, and z = number of 25 cent coins.

We know two things.

\(0.01w+0.05x+0.1y+0.25z=175\)

\(w+x+y+z=3576\)

Let's eliminate w because we can.

\(0.01(3576-x-y-z)+0.05x+0.1y+0.25z=175\)

\(0.04x+0.09y+0.24z=139.24\)

There will be more than one solution for this problem. Let's settle for finding just one of them by saying z=0 and y=0.

\(0.04x=139.24\)

\(x=3481\)

\(w+3481=3576\)

\(w=95\)

One solution to this problem would be 95 one cent coins, 3481 five cent coins 0 ten cent coins, 0 twenty-five cent coins.