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Delicious77 [7]

2 years ago

15

Prove that 1/sec0-tan0= sec0+tan0

1 answer:

Leviafan [203]2 years ago

7 0

Answer:

Step-by-step explanation:

$\text{LHS}=\frac{1}{\sec \theta-\tan \theta}\\=\frac{\sec \theta+\tan \theta}{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}\\=\frac{\sec \theta+\tan \theta}{\sec^{2} \theta-\tan^{2} \theta}\\=\frac{\sec \theta+\tan \theta}{1}\\=\sec \theta+\tan \theta\\=\text{RHS}$

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If you can just help me with one question it would help very very much:)

Soloha48 [4]

1) a.30
3) complementary: 12 and 78
supplementary: 102 and 78

6 0

3 years ago

5=2/3x what does x equal

mixer [17]

Answer:

I dont know if u mean 5 equals 2 divided by 3x, but I got 2/15

Step-by-step explanation:

3 0

3 years ago

Sal Boxer decided to divide a gift of $5000 into two different accounts. He placed $1000 in an account that earns an annual simp

mash [69]

Answer:

THE TOTAL INTEREST EARNED = $ 385

Step-by-step explanation:

Interest earned by sal from first account can be calculated as follows :

Principal value (P) = $1000 .

interest rate (R) = 7.5 % .

Time period (T) = 1 year .

$\textbf{S . I} =\frac{\texbf{P * R * T}}{\textbf{100}}$

$= \frac{1000 * 7.5 * 1}{100}$

= $ 75 .

SIMILARLY , we can find the intrest earned by rest amount.

Principal value (P) = $5000 - $ 1000 = $ 4000 .

Interest rate (R) = 7.75 % .

Time period (T) = 1 year .

$S . I = \frac{4000 * 7.75 * 1}{100}$

= $ 310 .

THE TOTAL INTEREST EARNED = $ 75 + $ 310

= $ 385

4 0

3 years ago

a) Does the Range Rule of Thumb estimate more often underestimate or overestimate the actual standard deviation?

qwelly [4]

Yes it dose hope this helped :)

8 0

3 years ago

Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well

laiz [17]

Answer:

The value is $P(A) = 0.133617$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 7.5$

The standard deviation is $\sigma = 0.2$

The safest water level is between 7.2 and 7.8

Generally the probability that the selected pool has a pH level that is not considered safe is mathematically represented as

$P(A) = 1 - P(7.2 \le X \le 7.8 )$

Here

$P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - \mu }{\sigma } < \frac{X - \mu }{ \sigma }$

Generally $\frac{X - \mu }{ \sigma } = Z (The \ standardized \ value \ of X )$

So

$P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - 7.5 }{0.2 } < Z$

$P(7.2 < X < 7.8 ) = P(-1.5 < Z$

=> $P(7.2 < X < 7.8 ) = P(Z < 1.5) - P( Z < - 1.5)$

From the z-table the probability of (Z < -1.5) and ( Z <1.5) are

$P(Z < 1.5) = 0.93319$

and

$P(Z < -1.5) = 0.066807$

So

$P(7.2 < X < 7.8 ) =0.93319 - 0.066807$

$P(7.2 < X < 7.8 ) =0.0866383$

So

$P(A) = 1 - 0.0866383$

=> $P(A) = 0.133617$

5 0

3 years ago