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3 years ago

Write log146−log14 3 2 as a single logarithm.

Mathematics

2 answers:

attashe74 [19]3 years ago

5 0

Answer:

the answer is log 14(4) on edg.

Step-by-step explanation:

Elden [556K]3 years ago

3 0

The answer to this is log14 4

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We say that an integer a is a type 0 integer if there exists an integer n such that a = 3n. An integer a is a type 1 integer if

Delicious77 [7]

Answer:

<em>Proof below</em>

Step-by-step explanation:

Let's assume a is a type 1 integer. By definition, it means we can find an integer n such that

a=3n+1

We need to prove $a^2$ is a type 1 integer

Expanding

$a^2=(3n+1)^2=9n^2+6n+1$

If $a^2$ is a type 1 integer, then we should be able to find an integer m such as

$a^2=3m+1$

Equating

$a^2=3m+1=9n^2+6n+1$

solving for m

$m=3n^2+2n$

Since we know n is an integer, then the expression of m gives an integer also. Having found the required integer m, the assumption is proven

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3 years ago

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Round 45728 to the nearest hundred

TiliK225 [7]

The answer is 45,700

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3 years ago

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2. Read "Lisa's Garden" and answer the question.

Ber [7]

38 Is it explain step by step

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2 years ago

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GIVING CROWN IS RIGHT!!!!!!

nasty-shy [4]

All you have to do is multiply the l * w = Area

When you look at 26 ft and 15 ft, 21 ft and 10 ft, 29 ft and 18 ft, you'd know it wouldn't work since it's over 180.

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3 years ago

Explain the error a student is asked to find when the value of an investment of $5200 is an account that earns 4.2% annual inter

zloy xaker [14]

We are given the following information

Investment = $5200

Annual interest rate = 4.2% = 0.042

Final amount = $16,500

Number of years = 27.7 years

Number of compoudings = quartely = 4

The student uses the following model

$V(t)=5200(1.014)^{3t}$

The general formula for compound interest is given by

$V(t)=P(1+\frac{r}{n})^{nt}$

As you can see, the number of compoundings is incorrect (3 vs 4)

The interest rate is also incorrect.

Let us substitute the given values into the above formula

$\begin{gathered} V(t)=P(1+\frac{r}{n})^{nt} \\ V(t)=5200(1+\frac{0.042}{4})^{4\cdot27.7} \\ V(t)=5200(1+0.0105)^{110.8} \\ V(t)=5200(1.0105)^{110.8} \\ V(t)=\$16543.5^{} \end{gathered}$

Therefore, the final amount is approximately $16,543.5

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1 year ago