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Dafna11 [192]
3 years ago
12

Which expression is equivalent to log c x2-1/5x

Mathematics
2 answers:
Aleks04 [339]3 years ago
6 0
Answer
C is the correct answer

㏒(x²-1) - (㏒5 + ㏒x)


Explanation
The law of logarithm says; 
㏒AB = ㏒₃A + ㏒₃B
㏒C/B = ㏒₃C - ㏒₃B
So, the given equation will be,

㏒((x²-1)/5x) = ㏒(x²-1) - ㏒5x
                      = ㏒(x²-1) - (㏒5 + ㏒x)

              
navik [9.2K]3 years ago
6 0
 Hello! For this problem we need to make use of two existing logarithmic properties. These are:

1. log \frac{M}{N} is equal to log M-logN
2. log(MN) is equal to log M+logN

Following the first property, we can simplify the expression to log (x^{2}-1)-log(5x).

Then, we will use the second property to the term 5x. The final form of the expression would then be log(x^{2}-1)-(log5+logx).

ANSWER: log(x^{2}-1)-(log5+logx)  (third option)
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Please help with this pleaseee
saw5 [17]

Answer:

CED=126

Step-by-step explanation:

The total of all 3 angles should be 180

so:

180-38-16 (ABC is equal to ECD)

180-38-16=126

126=CED

5 0
2 years ago
Read 2 more answers
Can someone help? This is hard
Licemer1 [7]

CT=90°

BTA=90°

x=29

CT=37

TA=20

CA=57

Step-by-step explanation:

they say BT is an altitude ( or height) . Altitudes are perpendicular meaning that:

BTC =90°

BTA=90°

Therefore:

(3x+3)°=90°

3x=90-3

3x=87°

x=29

CT:

x+8

=29+8

=37

TA:

x-9

=29-9

=20

CA:

=CT+TA

=37+20

=57

7 0
3 years ago
The time rate of change of a rabbit population PP is proportional to the square root of PP. At time t=0t=0 (months) the populati
frosja888 [35]

Answer:

\frac{dP}{\sqrt{P}} = k dt

And if we integrate both sides we got:

2 \sqrt{P} = kt +C

Where C is a constant., we can rewrite the expression like this:

\sqrt{P} = \frac{1}{2} (kt +C)

If we square both sides we got:

P = \frac{1}{4} (kt +C)^2

If we use the initial condition we have that:

P(0) = 100 = \frac{1}{4} (k*0 +C)^2

And we can solve for C like this:

400 = C^2

C = 20

And now we can find the derivate of the function and we got:

P'(t) = 2* \frac{1}{4} (kt + 20) * k

Using the condition P'(0) = 10 we got:

10 = \frac{1}{2} k (k*0 +20)

20 = 20 k

k= 1

And then the model is defined as:

P = \frac{1}{4} (t +20)^2

And for t =12 months we have:

P(12) = \frac{1}{4} (12 +20)^2 = 256

Step-by-step explanation:

For this case we cna use the proportional model given by:

\frac{dP}{dt} = k \sqrt{P}

Where k is a proportional constant, P the population and the represent the number of months

For this case we know the following initial condition P(0) =100 and P'(0) = 10

we can rewrite the differential equation like this:

\frac{dP}{\sqrt{P}} = k dt

And if we integrate both sides we got:

2 \sqrt{P} = kt +C

Where C is a constant., we can rewrite the expression like this:

\sqrt{P} = \frac{1}{2} (kt +C)

If we square both sides we got:

P = \frac{1}{4} (kt +C)^2

If we use the initial condition we have that:

P(0) = 100 = \frac{1}{4} (k*0 +C)^2

And we can solve for C like this:

400 = C^2

C = 20

And now we can find the derivate of the function and we got:

P'(t) = 2* \frac{1}{4} (kt + 20) * k

Using the condition P'(0) = 10 we got:

10 = \frac{1}{2} k (k*0 +20)

20 = 20 k

k= 1

And then the model is defined as:

P = \frac{1}{4} (t +20)^2

And for t =12 months we have:

P(12) = \frac{1}{4} (12 +20)^2 = 256

6 0
3 years ago
Earth is 149 597 890 km from the sun. Pluto is about 39.5 times farther away from the sun than earth. how far is pluto from the
leonid [27]

Answer:

39.5 astronomical units

4 0
3 years ago
Can someone help me with this
Alla [95]

Answer:

x=125°

Step-by-step explanation:

180-(25+30)= 125°

hope it helps!

7 0
3 years ago
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