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3 years ago

16,8,4,2,1 find the next two terms

Mathematics

2 answers:

11Alexandr11 [23.1K]3 years ago

3 0

The next terms are 1/2 and 1/4. you divide by two next each.

butalik [34]3 years ago

3 0

<h3>The next two terms are: 1/2 and 1/4</h3>

note: in decimal form, 1/2 = 0.5 and 1/4 = 0.25

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Explanation:

This sequence is geometric with the common ratio r = 1/2 = 0.5

Multiply each term by 1/2 to get the next

16 times 1/2 = 16 times 0.5 = 8

8 times 1/2 = 8 times 0.5 = 4

4 times 1/2 = 4 times 0.5 = 2

2 times 1/2 = 2 times 0.5 = 1

1 times 1/2 = 1/2 = 0.5

1/2 times 1/2 = 1/4 = 0.25

So that's how we end up with 1/2 = 0.5 and 1/4 = 0.25 as the next two terms.

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As a shortcut, you can use the nth term formula

an = 16*(1/2)^(n-1)

and plug in n = 6 and n = 7 to generate the next two terms a6 and a7. You should get 1/2 and 1/4 as your results.

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What is the next fraction in this sequence? Simplify your answer. 3/4, 1/4, 1/12, 1/36

antiseptic1488 [7]

Hey their,

3/4
1/4
1/12
1/36
0/60

The pattern is -1/x every 2 +x/8

How i got this was 36/12=3
3*12=36
so 60-36=24
24x3/2+24=60
then 60/5=12

Hopethis helped

7 0

4 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so