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aniked [119]
3 years ago
13

Find all the solutions in the interval [0,2π) sin²x = 2-2cosX

Mathematics
1 answer:
Finger [1]3 years ago
5 0

Step-by-step explanation:

sin²x = 2 - 2cosx

1 - cos²x = 2 - 2cosx (because sin²x + cos²x = 1)

cos²x - 2cosx + 1 = 0

(cosx - 1)² = 0

cosx - 1 = 0

cosx = 1

Hence x = 0.

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A TV is on sale at 30% off for $420. What was the original price of the TV?
liberstina [14]
<span>"A TV is on sale at 30% off for $420"
 
What was the original price of the TV?
</span>

Well if you start to calculate first off you are sooner or later going to figure out that in order for it to be 420 the answer has to be bigger than 1,000. If you get to 1,200 and calculate 30 percent off of it it the answer is 360 so a ballpark quesstiminate would be to go two more hundredths up. Therefore 30 percent of 1,400 is 420
4 0
3 years ago
Multiply the coefficients<br> and add the exponents<br> for the same variables
GREYUIT [131]
How do i answer this?
8 0
3 years ago
JK and LM are straight lines. Find /NOK.<br> ( help me pls)<br>​
VashaNatasha [74]

Answer:

NOK = 58

Step-by-step explanation:

Given

JOM = 108

LON = 50

Required

Find NOK

From the attached triangle, we have:

LOK = JOM =108 --- Corresponding angles

And

LOK = LON + NOK

Substitute for LOK and LON

108= 50+ NOK

Make NOK the subject

NOK = 108 - 50

NOK = 58

8 0
3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
When planning road development, the road commission estimates the future population using the function represented in the table,
Sindrei [870]

The significance of 160,000 in the function is the initial population at the time of the estimation.

The correct option is (C)

<h3>What is data?</h3>

A collection of facts, such as numbers, words, measurements, observations or even just descriptions of things.

Given:

As per the table attached below:

F(x) signify the total population when the time was x hour.

So,  160,000 in the function signifies that the population at starting i.e., initial point.

Learn more about this concept here:

brainly.com/question/17112729

#SPJ1

7 0
2 years ago
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