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RoseWind [281]
3 years ago
9

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)=

In(3x/0); division by 0, undefined
Mathematics
1 answer:
Art [367]3 years ago
4 0

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

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Answer:

Step-by-step explanation:

Let :

C = number of cats

D = number of dogs

Raul's pet store has a play area that can fit up to 30 cats and dogs.

C + D = 30

The pet store never has more than 8 cats in the play areas.

<h2>C < 9</h2>

(there can never be 9 or more cats)

As or the number of dogs :

C + D = 30

C = 30 - D

Since we know that C < 9. To get the number of dogs allowed, we just plug in 30 - D for C.

30 - D < 9

30 - 9 < D

or

<h2>D > 21</h2>

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5 0
1 year ago
The dot plot shows the number of classes per semester that a group of boys and girls have chosen to take in their freshman year
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Answer:

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Step-by-step explanation:

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\text{binominals:}\\\\C.\ x^2+18\\D.\ x+2
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60 mi. = _______________ km
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Step-by-step explanation:

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If j and k are nonzero integers, which pair of points must lie in the same quadrant?
Marizza181 [45]
<h3>Answer:  Choice D.   (3j, 3k)  and (3/j, 3/k)</h3>

The slash indicates a fraction.

=============================================

Proof:

We'll need to consider 4 different cases.

-----------------------

Case (1): j > 0 and k > 0

If j > 0, then 3j > 0 and 3/j > 0

If k > 0, then 3k > 0 and 3/k > 0

The two points (3j, 3k) and (3/j, 3/k) are both in quadrant 1.

-----------------------

Case (2): j > 0 and k < 0

If j > 0, then 3j > 0 and 3/j > 0

If k < 0, then 3k < 0 and 3/k < 0

Points (3j, 3k) and (3/j, 3/k) are both in quadrant 4.

------------------------

Case (3): j < 0 and k > 0

If j < 0, then 3j < 0 and 3/j < 0

If k > 0, then 3k > 0 and 3/k > 0

Points (3j, 3k) and (3/j, 3/k) are in quadrant 3.

------------------------

Case (4): j < 0 and k < 0

If j < 0, then 3j < 0 and 3/j < 0

If k < 0, then 3k < 0 and 3/k < 0

Points (3j, 3k) and (3/j, 3/k) are in quadrant 4.

------------------------

For nonzero integers j and k, we've shown that Points (3j, 3k) and (3/j, 3/k) are in the same quadrant. This concludes the proof.

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