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3 years ago

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Arrange the integers 1,4,8,12,17,32 so that all consecutive integers add to a perfect square. Enter your answer with numbers sep

arated by commas.

1 answer:

viktelen [127]3 years ago

6 0

One possible answer is: 1, 8, 17, 32, 4, 12

I don't know if there's a methodical way to get this answer. I used trial and error. I started with 1 and looked through the list to see what adds to 1 to get a perfect square. That would be 8 since 1+8 = 9 = 3^2

The process is repeated but this time for 8. After a bit of guess and checking, we see that 8+17 = 25 = 5^2.

Then after 17 is 32 because 17+32 = 49 = 7^2.

Keep doing this until all of the values are used up. If you get stuck, then try backtracking to a previous branch/path where everything worked and try another fork in the road. If that doesn't work, then try starting the sequence with a completely different value (instead of 1).

Other answers may be possible. I haven't checked all possible permutations.

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62 62 65 65 115 115 118 118<br> I have no idea how to solve this, will give brainliest

Kobotan [32]

Answer:

1.ABC = 45°,DBE = 45°. ABD = 135°,CBE = 135°

2.ABC = 62°, DBE = 62°, ABD =118°,CBE = 118°

Step-by-step explanation:

1.angle ABC = angle DBE because they are vertical. You can write the equation 3x+38 = 5x+20 and find both angles. When you solve, x=9. You substitute it in. ABD is equal to 180-45° because it is supplementary with ABC. ABD = CBE b/c vertical angles.

2.You do the same thing. 4x+2=5x-13,x=15. You substitute in 15 for x and get 62°. They are the equal angles. For ABD you do 180 -62 since they are supplementary angles, which gets you 118 which applies for both because they are vertical angles.

I want the brainliest.

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The upper quartile is 23

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What fraction is a multiple of1/9

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Find an equation for the line perpendicular to the line 8x - 6y = -2 having the same y-intercept as - 6x + 2y = 4.

MissTica

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3 x - y + 2 = 0

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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

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

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3 years ago