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3 years ago

Simplify log (x2-y2)- log (x-y)=

2 answers:

8 0

Using the law of logarithm. LogA - LogB = Log(A/B)

log(x²-y²) - log(x-y) = log((x²-y²)/(x-y))

Note by difference of two squares, (x²-y²) = (x-y)(x+y)

Simplifying (x²-y²)/(x-y) = (x-y)(x+y)/(x-y) = (x+y)

Therefore log(x²-y²) - log(x-y) = log((x²-y²)/(x-y)) = log((x-y)(x+y)/(x-y)) = log(x+y)

jok3333 [9.3K]3 years ago

3 0

Remember:
log A - log B=log (A/B)
(a²-b²)=(a+b)(a-b).

Therefore

log (x²-y²)-log(x-y)=log[(x²-y²)/(x-y)]= log[(x+y)(x-y) / (x-y)]=log (x+y)

Answer: log (x²-y²)-log(x-y)=log (x+y)

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a bag containing some widgets weighing 81 kg. when 2 of the widgets are removed, the weight of the bag drops to 73kg. the bag wi

Sloan [31]

First I find the individual widget weight

81 - 73 = 2x
8 = 2x
x = 4

Then I find how many widgets were left

73 - 1 = 4y
72 = 4y
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3 0

3 years ago

Put in simplest form 8/15 × 5/6

algol13

8/15 x 5/6
•multiply across the numerator and denominator.

= 40/90
• then find a common factor that will go into both the numerator and denominator equally.

Common factor: 5

= 8/18
•there is another common factor, which is 2

Simplified:

=4/9

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3 years ago

In 10 seconds, Lauren counts 5 vehicles passing by her house. At that rate, how many vehicles would pass by in one hour?

Anna007 [38]

6 x 60 x 5 would be the equation and all of that put together would be 1,800 vehicles would pass by in an hour.

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3 years ago

Now move the center of rotation, R, to a different position on the coordinate plane.

Scilla [17]

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3 years ago

Suppose that in a senior college class of 500500 students, it is found that 179179 smoke, 228228 drink alcoholic beverages, 1

olga2289 [7]

Answer: a) 0.16, b) 0.058, and c) 0.856.

Step-by-step explanation:

Since we have given that

Number of students = 500

Number of students smoke = 179

Number of students drink alcohol = 228

Number of students eat between meals = 119

Number of students eat between meals and drink alcohol = 59

Number of students eat between meals and smoke = 72

Number of students engage in all three = 30

a) Probability that the student smokes but does not drink alcohol is given by

$P(S-A)=P(S)-P(S\cap A)\\\\P(S-A)=\dfrac{179}{500}-\dfrac{99}{500}\\\\P(S-A)=\dfrac{179-99}{500}\\\\P(S-A)=\dfrac{80}{500}\\\\P(S-A)=0.16$

b) eats between meals and drink alcohol but does not smoke.

$P((M\cap A)-S)=P(M\cap A)-P(M\cap S\cap A)\\\\P((M\cap A)-S)=\dfrac{59}{500}-\dfrac{30}{500}\\\\P((M\cap A)-S)=\dfrac{59-30}{500}\\\\P((M\cap A)-S)=\dfrac{29}{500}\\\\P((M\cap A)-S)=0.058$

c) neither smokes nor eats between meals.

$P(S'\cap M')=1-P(S\cup M)\\\\P(S'\cap M')=1-\dfrac{72}{500}\\\\P(S'\cap M')=\dfrac{500-72}{500}\\\\P(S'\cap M')=\dfrac{428}{500}=0.856$

Hence, a) 0.16, b) 0.058, and c) 0.856.

5 0

3 years ago