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Sergio039 [100]

4 years ago

12

Someone answer this please

1 answer:

Mice21 [21]4 years ago

5 0

Answer:

Which one?

Step-by-step explanation:

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Sedbober [7]

If you're talking about unit cubes, then the equation is

length*width*height

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5. Which side lengths could not form a triangle?

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A, B, C, D

Step-by-step explanation:

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3 years ago

Find the error & find the correct answer 2In(x)=In(3x)-[In(9)-2In(3)] In(x^2)=In(3x)-[In(9)-In(9)] In(x^2)=In(3x)-0 In(x^2)=

Art [367]

Answer:

Error: $lnx^2=ln 3x$ not $ln\frac{3x}{0}$

Solution:x=0 and 3

Step-by-step explanation:

We have to find the error and correct answer

Given: $2ln x=ln(3x)-[ln9-2ln(3)]$

$lnx^2=ln(3x)-[ln9-ln3^2]$

Using the formula

$alog b=logb^a$

$lnx^2=ln(3x)-[ln9-ln9]$

$lnx^2=ln(3x)-0$

$lnx^2=ln(3x)$

$x^2=3x$

$x^2-3x=0$

$x(x-3)=0$

Therefore, x=0 and x=3

But last step in the given solution

$lnx^2=ln\frac{3x}{0}=\infty$

It is wrong this property is used when

$log m-log n$ then

$log\frac{m}{n}$

Hence, the student wrote $lnx^2=ln\frac{3x}{0}$ instead of $lnx^2=ln3x$ and solution is given by

x=0 and x=3

4 0

3 years ago

Find the average value of the function on the given interval.<br> f(x)=(2x-1)1/2; [1,13]

Leni [432]

Answer:

The average value of the function on the given interval 6.5.

Step-by-step explanation:

Consider the given function is

$f(x)=\dfrac{2x-1}{2}$

We need to find the average value of the function on the given interval [1,13].

$f(x)=\dfrac{2x}{2}-\dfrac{1}{2}$

$f(x)=x-0.5$

The average value of the function f(x) on [a,b] is

$Average=\dfrac{1}{b-a}\int\limits^b_a {f(x)} \, dx$

Average value of the function on the given interval [1,13] is

$Average=\dfrac{1}{13-1}\int\limits^{13}_{1} {x-0.5} \, dx$

$Average=\dfrac{1}{12}[\dfrac{x^2}{2}-0.5x]^{13}_{1}$

$Average=\dfrac{1}{12}[\dfrac{(13)^2}{2}-0.5(13)-(\dfrac{(1)^2}{2}-0.5(1))]$

$Average=\dfrac{1}{12}[78-0]$

$Average=6.5$

Therefore, the average value of the function on the given interval 6.5.

6 0

3 years ago