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3241004551 [841]
4 years ago
9

Complete the pattern.2.78, 4.03, 5.78, 8.03, 10.78,_____

Mathematics
1 answer:
babunello [35]4 years ago
4 0

The next number in the pattern would be 14.03.

To find this, note the differences found by subtracting each number from the number that comes after.

4.03 - 2.78 = 1.25

5.78 - 4.03 = 1.75

8.03 - 5.78 = 2.25

10.78 - 8.03 = 2.75

Since each difference goes up by .5 each time, we know we can add .5 to the last difference to find the newest one.

2.75 + .5 = 3.25

Now we have the difference, so we add that to the last number.

10.78 + 3.25 = 14.03

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on a cross-country road trip, a family drives 240 miles each day. Write and solve an equation to determine how far the family ha
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Well, first, let's label the variables for our equation:

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Identify the absolute extrema of the function and the x-values where they occur.
sladkih [1.3K]
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For x the derivative is negative.
For x>2 the derivative is positive.
Therefore at x=2 there's a minimum.

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Please help math questions <br> put number with answers please
natita [175]

Answer:

See below for answers

Step-by-step explanation:

<u>Problem 1</u>

<u />\frac{1}{sin\theta}=2cos\theta\: ; \: 0\leq\theta < 2\pi\\\\1=2sin\theta cos\theta\\\\1=sin2\theta\\\\\frac{\pi}{2}+2\pi n=2\theta\\ \\\frac{\pi}{4}+\pi n=\theta\\ \\\theta=\bigr\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\bigr\}

Therefore, the second option is correct.

<u>Problem 2</u>

<u />cos(2x)+1=sin(x)+2\:;\: 0\leq x < 2\pi\\\\1-2sin^2x+1=sin(x)+2\\\\-2sin^2x=sin(x)\\\\0=2sin^2x+sinx\\\\0=sinx(2sinx+1)

0=sinx\\\\x=\pi n\\\\x=0,\pi

0=2sinx+1\\\\-1=2sinx\\\\-\frac{1}{2}=sinx\\ \\x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\\\x=\frac{7\pi}{6},\frac{11\pi}{6}

Therefore, the solution set is x=\bigr\{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\bigr\}, making the second option correct.

<u>Problem 3</u>

<u />2cos^2x=-cosx\: ;\: 0^\circ\leq x < 360^\circ\\\\2cos^2x+cosx=0\\\\cosx(2cosx+1)=0

cosx=0\\\\x=\frac{\pi}{2}+\pi n\\ \\x=90^\circ+180n^\circ\\\\x=90^\circ,270^\circ

2cosx+1=0\\\\2cosx=-1\\\\cosx=-\frac{1}{2}\\\\x=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n\\ \\x=120^\circ+360n^\circ,240^\circ+360n^\circ\\\\x=120^\circ,240^\circ

Therefore, the solution set is x=\bigr\{90^\circ,120^\circ,240^\circ,270^\circ\bigr\}, making the fourth option correct

7 0
2 years ago
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