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tensa zangetsu [6.8K]
3 years ago
11

Find the interval in which f(x)=sinx−cosx is increasing or decreasing?

Mathematics
1 answer:
alisha [4.7K]3 years ago
3 0

Answer:

There is no short answer.

Step-by-step explanation:

To find to intervals which f(x) increases or decreases, we first need to find it's derivative.

f(x) = sinx - cosx\\f'(x) = cosx - (-sinx) = cosx + sinx

The function is increasing when it's value is  > 0 and decreasing when it's value is < 0.

If we take a look at this graph, cosx+sinx is positive when they are both positive or when cosx is greater then sinx on the negative part.

I hope this answer helps.

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A microwave was bought for 15,000 and sold for 14,000.What is the percentage loss?
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Answer:

~6.7%

Step-by-step explanation:

A microwave was bought for 15,000 and sold for 14,000

=> The loss is 15000 - 14000 = 1000

=> The percentage of loss is 1000/15000= 0.066 = ~6.7% of original price

Hope this helps!

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Oliver was on the internet researching prices of new and used cars. One website has 47 cars for sale, 15 of which were convertib
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Answer: It a permutation

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The graph shows function g, a transformation of f(x)= x^1/3
snow_tiger [21]
The answer is B !!??!
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Ms. Rose bought a pack of star stickers. Out of every 10 stars, 4 are gold. If there are 60 stars in the pack, what fraction of
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for the fraction its 2,5 for the percentage its 40%

3 0
3 years ago
7 measured data points have a sample mean of 1403 and a standard deviation of 27. Determine the best estimate of the mean value
mina [271]

Answer:

Step-by-step explanation:

Hello!

You have sample of n=7 with mean X[bar]= 1403 and standard deviation S=27 and are required to estimate the mean with a 95%CI.

Asuming this sample comes from a normal population I'll use a stuent t to estimate the interval (a sample of 7 units is too small for the standard normal to be accurate for the estimation):

[X[bar]±t_{n-1;1-\alpha /2} * \frac{S}{\sqrt{n} }]

t_{n-1;1-\alpha /2} = t_{6;0.975}= 2.365

[1403±2.365*\frac{27}{\sqrt{7} }]

[1378.87;1427.13]

The margin of error is the semiamplitude of the interval and you can calculate it as:

d= \frac{Upbond-Lowbond}{2}= \frac{1427.13-1378.87}{2} = 24.13

With a confidence level of 95% you'd expect that the real value of the mean is contained by the interval [1378.87;1427.13], the best estimate of the mean value is expected to be ± 24.13 of 1403.

I hope it helps!

4 0
3 years ago
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