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4 years ago

How do you simplify cscx*secx-tanx?

1 answer:

5 0

Csc x · sec x - tan x =
= 1 / sin x · 1 / cos x - sin x / cos x =
= 1 / sinx cos x - sin² x / sin x cos x =
= ( 1 - sin² x ) / (sin x cos x) =
= cos² x / ( sin x cos x ) =
= cos x / sin x = cot x

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John read the first 144 pages of a novel, which was 3 pages less than 1/3 of the novel. Write an equation to determine the total

igomit [66]

Answer: 441 pages

Step-by-step explanation:

$\frac{1}{3}p-3=144\\\frac{1}{3}p=147\\3(\frac{1}{3})p=147(3)\\p=441$

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3 years ago

5(3q-1)5q+15 <br><br> Please answer be fast

vodomira [7]

Answer:

15q

Step-by-step explanation:

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3 years ago

Read 2 more answers

Functionally important traits in animals tend to vary little from one individual to the next within populations, possibly becaus

Jet001 [13]

Answer:

The 95% confidence interval for the standard deviation is (0.14, 0.54).

Step-by-step explanation:

The complete data and question is:

Sloth CW Ratios ; {1.5 , 1.09 , 0.98 , 1.42 , 1.49 , 1.25}

The 95% confidence interval for the standard deviation of this data is < σ < (two decimals - include the leading zero) .

Solution:

Compute the sample standard deviation as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{6}\times 7.73=1.2883\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{6-1}\times 0.2387}=0.2185$

The (1 - <em>α</em>)% confidence interval for the variance is:

$CI=\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2}}$

Confidence level = 95%

⇒ α = 0.05

The degrees of freedom is,

df = n - 1 = 6 - 1 = 5

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.025,5}=12.833\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{0.975,5}=0.831$

*Use a Chi-square table.

Compute the 95% confidence interval for the variance as follows:

$CI=\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2}}$

$=\frac{5\times (0.2185)^{2}}{12.833}$

Thus, the 95% confidence interval for the standard deviation is (0.14, 0.54).

8 0

4 years ago

WILL GIVE BRAINLIEST! Please answer both questions

Sergio039 [100]

First one: 8x + 80

Second one: -8n + 26 (A)

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3 years ago

For 4 weeks in June, Cameron biked 3 1/4 miles each week and swam 2 1/2 miles each week. For 3 weeks in July, he biked 4 3/4 mil

avanturin [10]

Step-by-step explanation:

Given that,

For 4 weeks in June, Cameron biked 3 1/4 miles each week and swam 2 1/2 miles each week.

For 3 weeks in July, he biked 4 3/4 miles each week and swam 3 1/2 miles each week.

We need to find how much greater was the total distance Cameron bike and swim in July compared to the total distance he bike in swim in June.

In July, total distance is ,

$4\dfrac{3}{4}\times 3+3\dfrac{1}{2}\times 3\\\\= \dfrac{19}{4}\times 3+\dfrac{7}{2}\times 3\\\\=24.75\ \text{miles}$

In June, total distance is :

$=4\times 3\dfrac{1}{4}+4\times 2\dfrac{1}{2}\\\\=\dfrac{13}{4}\times 4+\dfrac{5}{2}\times 4\\\\=23\ \text{miles}$

Hence, in June total distance is 23 miles and in July total distance is 24.75 miles.

7 0

3 years ago