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2 years ago

Is 16.45 greater than, less than or equal to 16.454

Mathematics

2 answers:

MArishka [77]2 years ago

8 0

we know that

$16.454=16.45+0.004$

$16.45 < 16.45+0.004$

therefore

$16.45$ is less than $16.454$

because $16.454$ is $4$ thousandths more than $16.45$

the answer is

$16.45$ is less than $16.454$

vampirchik [111]2 years ago

4 0

16.45 is less than 16.454. The reason is because 16.454 is 4 thousandths more than 16.45 assuming that both numbers are exact numbers. Although it is only a small amount it still makes 16.45 less than 16.454.

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105,159 rounded to the nearest ten thousand

ivolga24 [154]

For this case we have the following number:

105,159

By definition we have:

thousand place: five-digit number greater than zero.

On the other hand we have as a rule:

When the previous number is greater than or equal to five, then the next number increases by one.

So we have to round off to the nearest ten thousand:

105,159 = 110,000

Answer:

105,159 rounded to the nearest ten thousand is:

105,159 = 110,000

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Which of the following shows the extraneous solution to the logarithmic equation? log Subscript 4 Baseline (x) + log Subscript 4

vekshin1

Given:

The given equation is $\log _{4}(x)+\log _{4}(x-3)=\log _{4}(-7 x+21)$

We need to determine the extraneous solution of the equation.

Solving the equation:

To determine the extraneous solution, we shall first solve the given equation.

Applying the log rule $\log _{c}(a)+\log _{c}(b)=\log _{c}(a b)$ , we get;

$\log _{4}(x(x-3))=\log _{4}(-7 x+21)$

Again applying the log rule, if $\log _{b}(f(x))=\log _{b}(g(x))$ then $f(x)=g(x)$

Thus, we have;

$x(x-3)=-7 x+21$

Simplifying the equation, we get;

$x^2-3x=-7 x+21$

$x^2+4x=21$

$x^2+4x-21=0$

Factoring the equation, we get;

$(x-3)(x+7)=0$

Thus, the solutions are $x=3, x=-7$

Extraneous solutions:

The extraneous solutions are the solutions that does not work in the original equation.

Now, to determine the extraneous solution, let us substitute x = 3 and x = -7 in the original equation.

Thus, we get;

$\log _{4}(3)+\log _{4}(3-3)=\log _{4}(-7 \cdot 3+21)$

$\log _{4}(3)+\log _{4}(0)=\log _{4}(0)$

Since, we know that $\log _{a}(0)$ is undefined.

Thus, we get;

Undefined = Undefined

This is false.

Thus, the solution x = 3 does not work in the original equation.

Hence, x = 3 is an extraneous solution.

Similarly, substituting x = -7, in the original equation. Thus, we get;

$\log _{4}(-7)+\log _{4}(-7-3)=\log _{4}(-7(-7)+21)$

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$\log _{4}(-7)+\log _{4}(-10)=\log _{4}(70)$

Simplifying, we get;

Undefined = $\log _{4}(70)$

Undefined = 3.06

This is false.

Thus, the solution x = -7 does not work in the original solution.

Hence, x = -7 is an extraneous solution.

Therefore, the extraneous solutions are x = 3 and x = -7

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