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3 years ago

14

Determine whether each sequence is arithmetic, geometric, or neither. Then find a formula for the nTh term of the sequence 3,5,9

,17...

1 answer:

yan [13]3 years ago

3 0

Answer:

Neither

Step-by-step explanation:

<u>Given sequence:</u>

3, 5, 9, 17, ...

<u>We can write the terms as:</u>

3 = 2^1 + 1
5 = 2^2 + 1
9 = 2^3 + 1
17 = 2^4 + 1
...
aₙ = 2^n + 1

As we see the sequence doesn't meet AP or GP criteria, there is no common difference or common factor between the terms

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The box plot shows the heights of grade 7 students in two random samples from two different schools. The sample from each school

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The true statement is Grade 7 students from School A are typically shorter than grade 7 students from School B because of the difference in the medians of grade 7 student heights at the schools. (option D)

<h3>What is the true statement?</h3>

A box plot is used to study the distribution and level of a set of scores. The box plot consists of two lines and a box. the two lines are known as whiskers.

The end of the first line represents the minimum number and the end of the second line represents the maximum number.

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What is the product? 6(x^2-1) times 6x-1 over 6(x+1)

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(36x^3-36x-1)/6x+6 or 6x^2-6x+(-1/6x+6)

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3 years ago

Read 2 more answers

Write the quadratic equation that has roots (√3+1)/2 and (√3 -1 )/2, if its coefficient with x^2 is equal to:

NISA [10]

Answer:

The equation is: $x^2-x+\frac{1}{2}$

(A) $x^2-x+\frac{1}{2}=0$

(B) $5x^2-5x+\frac{5}{2}=0$

(C) $-\frac{1}{2}$

$-\frac{1}{2}x^2+\frac{1}{2}x-\frac{1}{4}=0$

d) √3

$(\sqrt{3}) x^2-x\sqrt{3}+\frac{\sqrt{3}}{2}=0$

Step-by-step explanation:

If the roots of the quadratic equation are: $\frac{\sqrt{3}+1}{2} and \frac{\sqrt{3}-1}{2}$

This means:

x= $\frac{\sqrt{3}+1}{2}$ OR $x= \frac{\sqrt{3}-1}{2}$

x- $x-\frac{\sqrt{3}+1}{2}=0 \: or\: x- \frac{\sqrt{3}-1}{2}=0$ =0 or x-

If a=0 or b=0, then ab=0

Therefore:

$(x-\frac{\sqrt{3}+1}{2})(x-\frac{\sqrt{3}-1}{2})=0\\x^2-\frac{x(\sqrt{3}-1)}{2}-\frac{x(\sqrt{3}+1)}{2}+(\frac{\sqrt{3}-1)}{2})(\frac{\sqrt{3}+1}{2})=0\\x^2-x+\frac{1}{2}=0$

Therefore the coefficient of $x^2$ here is 1.

b) To make 5 the coefficient of $x^2$

Simply multiply $x^2-x+\frac{1}{2}=0$ by 5

This gives: $5x^2-5x+\frac{5}{2}=0$

c) $-\frac{1}{2}$

To make $-\frac{1}{2}$ the coefficient of $x^2$

Simply multiply $x^2-x+\frac{1}{2}=0$ by $-\frac{1}{2}$

This gives: $-\frac{1}{2}x^2+\frac{1}{2}x-\frac{1}{4}=0$

d) √3

To make $\sqrt{3}$ the coefficient of $x^2$

Simply multiply $x^2-x+\frac{1}{2}=0$ by $\sqrt{3}$

This gives: $(\sqrt{3}) x^2-x\sqrt{3}+\frac{\sqrt{3}}{2}=0$

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4 years ago

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weeeeeb [17]

Answer:

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Step-by-step explanation:

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