Log 5 plus log 2 equals?

log5 + log2 =

Answers

Answer 1

Answer:
The sum of logs = the log of the product .

   log(5) + log(2) = log(5·2) = log(10)

If these are all logs to the base 10, then log(10) = 1

   log(5) + log(2) = 1

Let's try it out !

According to my calculator ...

   log(5) = 0.69897...

   log(2) = 0.30103...

Log(5) + log(2) = 0.99999... yay ! not bad

Answer 2

Answer:

The solution of the equation log5 + log2 using the properties of Logarithm is log(10) or 1.

Logarithms are used to solve exponential equations by converting them into simpler forms.

A number that represents the power to which a fixed number (the base) must be increased in order to get a specific number is called as logarithm.

The property of logarithms states that:

log(a) + log(b) = log(ab).

Applying this property, we get:

log5 + log2

= $\log(5*2)$

= log(10)

Logarithm base 10 (log) gives the power to which 10 must be raised to obtain the given value. So, log(10) represents the exponent required to obtain 10, which is 1.

Thus, the solution of log5 + log2 is equal to log(10) or 1.

Learn more about Logarithm here:

brainly.com/question/32351461

#SPJ6

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A company’s sales of garden tractors increased about 106% from 2007 to 2008. Th e number of tractors sold in 2007 was
347. How many tractors were sold in 2008?

Answers

Answer:

715 garden tractors.

Step-by-step explanation:

We have been given that a company’s sales of garden tractors increased about 106% from 2007 to 2008. The number of tractors sold in 2007 was 347.

So the number of tractors sold during 2008 will be 347 plus 106% of 347.

$\text{Number of tractors sold during 2008}=\text{Number of tractors sold in year 2007+ Increase in number of tractors in 2008}$

$\text{Number of tractors sold during 2008}=347+((106)/(100)* 347)$

$\text{Number of tractors sold during 2008}=347+(1.06* 347)$

$\text{Number of tractors sold during 2008}=347+367.82$

$\text{Number of tractors sold during 2008}=714.82\approx 715$

Therefore, 715 garden tractors were sold in 2008.

1) What is the equation of the line whose y-intercept is 3 and slope is 1?y = x - 3
y = x + 3
y = 3x + 1

Answers

$The\ slope-intercept\ form:y=mx+b\n\nm\ is\ slope\nb\ is\ y-intercept\n\nm=1;\ b=3\ then\ y=1x+3\Rightarrow y=x+3$

$y-intercept : \ b= 3 \n slope : \ m=1 \n \n the \ slope \ intercept \ form \ is : \n \n y= mx +b \n \n y=1\cdot x + 3 \n \ny=x+3$

Write as a product: a-b+a^2-B^2

Answers

Answer: (a - b)(a + b + 1)

a - b + a² - b²

= (a - b) + (a - b)(a + b)

= (a - b)(a + b + 1)

Step-by-step explanation:

Which pair of angles listed below are adjacent angles?Parallel Lines and Transversals

A.
1 and 2

B.
1 and 3

C.
1 and 4

D.
2 and 3

Answers

Lets look at the question is asking:

They are looking for adjacent angles.

The geometrical definition for adjacent is neighboring.

So with this we can tell that They are looking for neighboring angles.

The adjacent/neighboring angles are

∠1 and∠2 and ∠3 and ∠4.

So since ∠3 and∠4 are not an option, we can make an educated guess that the answer is A.

(I feel slightly confused on this question, so if it wrong I am truly sorry!)

I hope this helped :)

This math my life depends on it

Answers

A geometric sequence is one in which consecutive terms form a fixed ratio r. In other words, if aₙ is the nth term in the sequence, then

$a_n = a_(n-1)r$

For example, if a₁ = a is the 1st term, then

2nd term = a₂ = a₁r

3rd term = a₃ = a₂r = a₁r²

4th term = a₄ = a₃r = a₁r³

and so on. It's fairly easy to infer that

nth term = $a_n = a_(n-1)r = a_(n-2)r^2 = a_(n-3)r^3 = \cdots = a_1r^(n-1)$

14. We're given the 2nd and 5th terms, a₂ = -243 and a₅ = -9, and we use them to find the ratio r.

a₅ = a₄r = a₃r² = a₂r³

-9 = -243 r³

r³ = 1/27

⇒ r = 1/3

Then the 1st term is

a₁ = a₂/r = -243/(1/3) = -729

and the nth term is recursively given by

$a_n = \frac13a_(n-1)$

and explicitly by

$a_n = \left(\frac13\right)^(n-1) a_1 = -(729)/(3^(n-1)) = -(3^6)/(3^(n-1)) = -3^(7-n)$

15. Now we have a₄ = 1/72 and a₃ = -1/12. Using what we know about geometric sequences, we have

a₄ / a₃ = (a₃r) / a₃ = r

so that

r = (1/72) / (-1/12) = -1/6

Then the 1st term is

a₁ = a₂/r = a₃/r² = (-1/12) / (-1/6)² = -3

and the nth term is recursively given by

$a_n = -\frac16a_(n-1)$

and explicitly by

$a_n = \left(-\frac16\right)^(n-1) (-3) = -3\cdot(-6)^(1-n) = 3\cdot(-1)^n\cdot6^(1-n)$

Which term can be added to the list so that the greatest common factor of the three terms is 12h3?36h3, 12h6, __________

6h3

12h2

30h4

48h5

Answers

The term that can be added to the list so the GCF is 12h3 would be 48h5.

The reason being is that 48 is first divisible by 12 and does not yield a fraction, and we can remove upon dividing 3 h's from this term as it contains a total of 5 h's.

The term that can be added to the list so that the greatest common factor of the three terms is 12h3 is D. 48h5