gibsonj338

First take $17.60 and divide it by the number of slices to get what each slice costs.

$17.60/16 = $1.10

Multiply the cost of one slice by the number of slices eaten

$1.10 x 6 = $6.60

Brad ate $6.60 of the pizza.

This may be the right answer; however, the instructions say to use logarithmic differentation.  If anyone knows how to solve for the answer usin logarithmic differentation, I would really appreciate it.  Thanks.

Whoops.  I forgot to put to the third power on (t^4-1).  The equation should be h(t)=(t^4-1)^3(t^3+1)^4.  Can you please do it again?  I am so sorry.

Thanks for the help.  I way over thought this problem. y=mx+b makes total sense.  I feel really dumb.  Now I can finish my homework.  I do know how to use Desmos.

Huh?  What kind of an answer is that?  If you do not know the answer to these questions, then do not answer.

\(x+\frac{5}{x}-1+x+\frac{8}{x}+1=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{5}{x}-1+\frac{8}{x}+1=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{5}{x}-0+\frac{8}{x}=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{5}{x}+\frac{8}{x}=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{13}{x}=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{13}{x}-8x=8x+\frac{9}{{x}^{2}}-1-8x\)

\(-6x+\frac{13}{x}=8x+\frac{9}{{x}^{2}}-1-8x\)

\(-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1-0x\)

\(-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1-0\)

\(-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=\frac{9}{{x}^{2}}-1-\frac{9}{{x}^{2}}\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=\frac{0}{{x}^{2}}-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=0-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1=-1+1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1=0\)

\({x}^{2}(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1)=0\times{x}^{2}\)

\(-6{x}^{3}+\frac{13{x}^{2}}{x}-\frac{9{x}^{2}}{{x}^{2}}+1{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-\frac{9{x}^{2}}{{x}^{2}}+1{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-9+1{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-9+{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-9+{x}^{2}=0\)

\(-6{x}^{3}+{x}^{2}+13x-9=0\)

Since you cannot factor any further, the only way I know of to finish solving this equation is by graphing.

\({2}^{5}\times{4}^{\frac{1}{2}}\)

\(2\times2\times2\times2\times2\times\sqrt{{4}^{1}}\)

\(4\times2\times2\times2\times\sqrt{{4}^{1}}\)

\(8\times2\times2\times\sqrt{{4}^{1}}\)

\(16\times2\times\sqrt{{4}^{1}}\)

\(32\times\sqrt{{4}^{1}}\)

\(32\times\sqrt{4}\)

\(32\times2\)

\(64\)

Let x = phone's original price

Let 0.2 (20%) = discount of phone

Let 640 = price of phone after discount

\(0.2x=x-640\)

\(0.2x-x=x-640-x\)

\(-0.8x=x-640-x\)

\(-0.8x=-640-0\)

\(-0.8x=-640\)

\(\frac{-0.8x}{-0.8}=\frac{-640}{-0.8}\)

\(1x=\frac{-640}{-0.8}\)

\(x=\frac{-640}{-0.8}\)

\(x=800\)

\(59={17}^{x}\)

\({log}_{17}59=x\)

\(x=\frac{log59}{log17}\)

\(x≈1.4391918111143888\)

\(y=(x+h)2\)

\(y=2(x+h)\)

\(y=2x+2h\)

First find a point on the graph and replace \(x\) and \(y\) for the point.  The point I will use is the point \((2,0)\)

\(y=2x+2h\)

\(0=2(2)+2h\)

Next solve for \(h\).

\(0=2(2)+2h\)

\(0=4+2h\)

\(0-4=4+2h-4\)

\(-4=4+2h-4\)

\(-4=2h-0\)

\(-4=2h\)

\(\frac{-4}{2}=\frac{2h}{2}\)

\(-\frac{4}{2}=\frac{2h}{2}\)

\(-\frac{2}{1}=\frac{2h}{2}\)

\(-2=\frac{2h}{2}\)

\(-2=\frac{1h}{1}\)

\(-2=1h\)

\(-2=h\)

\(h=-2\)

The answer is \(h=-2\).  The equation is \(y=2(x+(-2))\) or \(y=2(x-2)\) or \(y=2x-4\)

\((x-9)({x}^{7}+{x}^{6}+{x}^{5}+{x}^{4}+{x}^{3}+{x}^{2}+x+1)\)

\({x}^{8}+{x}^{7}+{x}^{6}+{x}^{5}+{x}^{4}+{x}^{3}+{x}^{2}+1x-9{x}^{7}-9{x}^{6}-9{x}^{5}-9{x}^{4}-9{x}^{3}-9{x}^{2}-9x-9\)

\({x}^{8}-8{x}^{7}-8{x}^{6}-8{x}^{5}-8{x}^{4}-8{x}^{3}-8{x}^{2}-8x-9\)

Actually the answer is 899.9.