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3 years ago

Which is the simplest radical form of √52?

1 answer:

4 0

Short answer: 2 * sqrt(13)
Remark
There are a number of ways to look at this. I'll pick the easiest.

Step one
Factor 52 until there are no more prime factors to be used.

52 = 2 * 26
52 = 2 * 2 * 13. That's as far as you can go.

Rule
For every 2 equal prime factors, 1 of them can be taken out side of the root sign. The other one disappears.

sqrt(52) = sqrt(2*2* 13) = 2*sqrt(13)

Answer 2sqrt(13) <<<< answer

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3 years ago

State whether each sequence is arithmetic and justify your answer. If the sequence is arithmetic, write a recursive and an expli

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Answer:

Part A

$f(n)=52-12(n-1)$

$f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.$

Part B

$(2,4,8,16,32)\: \:$ Geometric Sequence

Part C

1/4,3/4,5/4,7/4,9/4

$g(n)=\frac{1}{4}+\frac{2}{4}(n-1)\\f(n)=\left\{\begin{matrix}1/4if \: \:n=1 & \\ f(n+1)+2/4& if\: n\geq 2 \end{matrix}\right$

Part D:

$h(n)=1.1+0.4(n-1)\\h(n)=\left\{\begin{matrix}1.1 & if\:n=1 \\ h(n+1)+0.4 & if\:n\geq 2\end{matrix}\right$

Step-by-step explanation:

By definition, an Arithmetic Sequence holds the same difference between each following number.

Part A

$(52,40, 28, 16)\\52-40=12\\40-28=12\\28-16=12\\d=12$

Explicit Formula

To write an explicit formula is to write it as function.

$f(n)=52-12(n-1)$

Recursive Formula

To write it as recursive formula, is to write it as recurrence given to some restrictions:

$f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.$

Part B

$(2,4,8,16,32)\: \:$

Geometric Sequence, since 2*2=4 8*2=16 and 16*2=32 and 8+2=10 8+16=24

Part C

$(\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4})\\\$

Arithmetic Sequence, difference

$d=\frac{2}{4}$

Explicit Formula:

$g(n)=\frac{1}{4}+\frac{2}{4}(n-1)$

Recursive Formula

$g(n)=\left\{\begin{matrix}\frac{1}{4} &if\:n=1 \\ g(n+1)+\frac{2}{4} &if\: n\geq 2\end{matrix}\right.$

Part D

(1.1,1.5,1.9,2.3,2.7) Arithmetic Sequence, difference d=0.4

Explicit formula

$h(n)=1.1+0.4(n-1)\\$

Recursive Formula

$h(n)=\left\{\begin{matrix}1.1 &if\:n=1 \\ h(n+1)+0.4 &if\: n\geq 2\end{matrix}\right.$

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