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DaniilM [7]
3 years ago
7

6 more than twice a number is greater than negative sixteen

Mathematics
1 answer:
Alexxandr [17]3 years ago
3 0

Writing the problem as an equation you have:

2x+6 > -16

Solve for x:


2x+6 > -16

Subtract 6 from each side:

2x > -22

Divide both sides by 2:

x > -11


The answer is: x > -11


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Ms. Smith has 50 pieces of candy. She gives Joe 30% of her candy. How many pieces does he get?
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15 pieces :)

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Step 2: -26.7 · -3

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Alenkasestr [34]

Answer:

x=\frac{4*(2+e)}{e-2}

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

log(\frac{1}{x} )=-log(x)

log(x*y)=log(x)+log(y)

Therefore:

log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1

Now, cancel logarithms by taking exp of both sides:

e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e

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2x^{2} -32=16e-8ex+ex^{2}

Substract 16e-8ex+ex^{2} from both sides and factoring:

-(x-4)*(-8-4e-2x+ex)=0

Multiply both sides by -1:

(x-4)*(-8-4e-2x+ex)=0

Split into two equations:

x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0

Solving for x-4=0

Add 4 to both sides:

x=4

Solving for -8-4e-2x+ex=0

Collect in terms of x and add 4e+8 to both sides:

x(e-2)=4e+8

Divide both sides by e-2:

x=\frac{4*(2+e)}{e-2}

The solutions are:

x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}

If we evaluate x=4 in the original equation:

log(0)-log(0)=1

This is an absurd because log (x) is undefined for x\leq 0

If we evaluate x=\frac{4*(2+e)}{e-2} in the original equation:

log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1

Which is correct, therefore the solution is:

x=\frac{4*(2+e)}{e-2}

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\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\stackrel{California}{\textit{14\% of 158609}}}{\left( \cfrac{14}{100} \right)158609}\implies 22205.26

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