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3 years ago

Math help! PLEAAASSEE

Mathematics

1 answer:

mina [271]3 years ago

5 0

The formula for the sum of a geometric sequence with first term "a", common ratio "r" and number of terms "n" is
S = a·(r^n -1)/(r -1)
You have a=2, r=-2, n=7, so the sum is
S = 2·((-2)^7 -1)/(-2 -1)
= 2(-128-1)/(-3)
= 2(43) = 86

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3(3u-2)=2(3u+3) What is the steps

sattari [20]

Hopes this helps:

Answer: u = 4

1. Expand.
9u - 6 = 6u + 6

2. Add 6 to both sides.
9u = 6u + 6 + 6

3. Simplify 6u + 6 + 6 to 6u + 12.
9u = 6u + 12

4. Subtract 6u from both sides.
9u - 6u = 12

5. Simplify 9u - 6u to 3u.
3u = 12

6. Divide both sides by 3.
u = 12/3

7. Simplify 12/3 to 4.
u = 4

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3 years ago

A consulting company must hire 20 new associates per year to replace those who have left the company for other positions or have

Murljashka [212]

Answer:

5.85 years

Step-by-step explanation:

Overall number of associates employed = 117

Number of associates employed per year = 20

Average = Total number of associates employed ÷ number of associates employed per year = 117 associates ÷ 20 associates per year = 5.85 years

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4 years ago

Read 2 more answers

What is 135 km/h to ft/s.

Dominik [7]

Answer:

123.035ft/s

Step-by-step explanation:

for an approximate result divide the speed value by 1.097

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3 years ago

If the area of the circle below is 18 ft2, what is the area of the shaded sector?

LuckyWell [14K]

Answer:

It would be C

Step-by-step explanation:

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3 years ago

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Please help math questions put number with answers please

natita [175]

Answer:

See below for answers

Step-by-step explanation:

Problem 1

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

Therefore, the second option is correct.

Problem 2

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

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

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

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

Problem 3

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

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

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

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