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stepladder [879]
3 years ago
12

Please help me sec theta - cos theta= sin^2 theta/ cos theta​

Mathematics
2 answers:
inn [45]3 years ago
6 0

Answer:

Taking LHS

=1 by cos theta- cos theta

= (1- cos²0)/ cos0

= sin²0/ cos0 ( because 1-cos²0 is also equals to sin²0)

=RHS hence proved

kotykmax [81]3 years ago
3 0

Answer:

Step-by-step explanation:

Let theta be β.

So,

sec\beta - cos\beta = \frac{sin^2\beta }{cos\beta } \\=>\frac{1}{cos\beta } - cos\beta = \frac{1 - cos^2\beta }{cos\beta } \\=>\frac{1 - cos^2}{cos\beta } = \frac{1 - cos^2\beta }{cos\beta }

Here , identity used = sin^2\beta + cos^2\beta = 1

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You're baking a cake with a ratio of 3:4. If you use 2 cups of flour,how many cups of sugar will you use?
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Answer:

2 2/3 cups of sugar

Step-by-step explanation:

If your ratio is 3 flour to 4 sugar, and you use 2 cups of flour, set up your proportion as follows:

\frac{3}{4}=\frac{2}{x}

Cross multiply to get 3x = 8 and x = 8/3 which is 2 and 2/3 cups of sugar.

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3 years ago
Find domain of this function. What you see are two rays on the graph.:
taurus [48]
The correct answer is C
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You pick a card at random. 4 5 6 7 What is P(less than 7)? Write your answer as a fraction or whole number.
marta [7]

Answer:

3/4 is less than 7.

Step-by-step explanation:

There are 4 numbers but there are only 3 less than 7.

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3 years ago
Cathy bought a bag of candy that weighed 0.74 pounds. What is this decimal as a fraction, in lowest terms?
Luden [163]

Answer:

Step-by-step explanation:

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3 years ago
Read 2 more answers
A random sample of 250 students at a university finds that these students take a mean of 15.3 credit hours per quarter with a st
SpyIntel [72]

Answer:  15.3\pm0.198

OR

(15.102, 15.498)

Step-by-step explanation:

The formula to find the confidence interval(\mu) is given by :-

\overline{x}\pm z_{\alpha/2, df}\dfrac{s}{\sqrt{n}}

, where n is the sample size

s = sample standard deviation.

\overline{x}= Sample mean

z_{\alpha/2} = Two tailed z-value for significance level of \alpha .

Given : Confidence level = 95% = 0.95

Significance level = \alpha=1-0.95=0.05

s= 1.6

\overline{x}=15.3

sample size : n= 250 , which is extremely large ( than n=30) .

So we assume sample standard deviation is the population standard deviation.

thus , \sigma=1.6

By standard normal  distribution table ,

Two tailed z-value for Significance level of 0.05 :

z_{\alpha/2}=z_{0.025}=1.96

Then, the 95% confidence interval for the mean credit hours taken by a student each quarter :-

15.3\pm (1.96)\dfrac{1.6}{\sqrt{250}}\\\\ =15.3\pm 0.19833\\\\=\approx15.3\pm0.198\\\\=(15.3-0.198,\ 15.3+0.198)=(15.102,\ 15.498)

Hence, the mean credit hours taken by a student each quarter using a 95% confidence interval. =(15.102, 15.498)

5 0
3 years ago
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