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Gala2k [10]
3 years ago
5

which number has one digit that represents 10 times the value of the digit to it's right? 375,595, 545,150 and 378,658

Mathematics
2 answers:
irga5000 [103]3 years ago
7 0

Answer:

The required number is : 375,595

Step-by-step explanation:

The given numbers are : 375,595, 545,150 and 378,658

We have to tell the number that has one digit that represents 10 times the value of the digit to it's right.

For this we need to look for the number that has two same digits together, because only then the left digit will be 10 times the right digit.

Only the number 375,595 has two 5's together where the left one is 10 times the right one.

Artyom0805 [142]3 years ago
3 0

It would be 375,595.

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Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1
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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

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Add 4 to both sides:

x=4

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

x(e-2)=4e+8

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

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log(0)-log(0)=1

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

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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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