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Find the exact value of tan A in simplest radical form

AlladinOne [14]

Answer:

12.68 de

Step-by-step explanation:

8 0

3 years ago

Integrate 1/((x^5)*sqrt(9*x^2-1)) ...?

Korolek [52]

Let 9x^2-1 = y^2
=> 18xdx = 2ydy 
=> ydy = 9xdx 

lower limit = sqrt(9*2/9 - 1) = sqrt(1) = 1 
upper limit = sqrt(9*4/9 - 1) = sqrt(3) 

Int. [sqrt(2)/3,2/3] 1/(x^5(sqrt(9x^2-1)) dx 

= Int. [sqrt(2)/3,2/3] xdx/(x^6(sqrt(9x^2-1)) 
= 81* Int. [1,sqrt(3)] ydy/((y^2+1)^3y) 
=81* Int. [1,sqrt(3)] dy/(y^2+1)^3 

y=tanz 
dy = sec^2z dz 

=81*Int [pi/4,pi/3] cos^4(z) dz 
=81/4*int [pi/4,pi/3] (1+cos(2z))^2 dz 
=81/4* Int. [pi/4,pi/3] (1+2cos(2z)+cos^2(2z)) dz 

=81/4*(pi/3-pi/4) + 81/4*(sin(2pi/3)-sin(pi/2)) + 81/8 * (pi/3-pi/4) 
+ 81/32 *(sin(-pi/3)-sin(pi)) 
=81(pi/48+pi/96+1/4*(sqrt(3)/2 - 1) - 1/32 * sqrt(3)/2) 
=81/32*(pi+3sqrt(3)-8)

5 0

3 years ago

30 Points

Zina [86]

Answer:

C. a property of algebra

Step-by-step explanation:

I hope this helps

4 0

3 years ago

Today only, a sofa is being sold at a 32% discount. The sale price is $493.

Naddik [55]

Answer:

yesterday the price was. $650.76

8 0

3 years ago

A single die is rolled twice. The set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),

lana [24]

Answer:

(9) $\frac{1}{12}$ (10) $\frac{1}{12}$ (11) $\frac{5}{12}$ (12) $\frac{1}{4}$ (13) $\frac{1}{6}$ 14) $\frac{5}{36}$ (15) $\frac{1}{12}$ (16)0

Step-by-step explanation:

The sample Space of the single die rolled twice is presented below:

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),}.

n(S)=36

(9)Probability of getting two numbers whose sum is 9.

The possible outcomes are: (3, 6), (4, 5), (5, 4)

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

10) Probability of getting two numbers whose sum is 4.

The possible outcomes are: (1, 3),(2, 2),(3, 1),

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

11.)Find the probability of getting two numbers whose sum is less than 7.

The possible outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)

$P(\text{two numbers whose sum is less than 7})=\frac{15}{36}=\frac{5}{12}$

12.Probability of getting two numbers whose sum is greater than 8

The possible outcomes are:(4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)

$P(\text{two numbers whose sum is greater than 8})=\frac{9}{36}=\frac{1}{4}$

(13)Probability of getting two numbers that are the same (doubles).

The possible outcomes are:(1, 1)(2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

$P(\text{two numbers that are the same})=\frac{6}{36}=\frac{1}{6}$

14.Probability of getting a sum of 7 given that one of the numbers is odd.

The possible outcomes are: (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

$P(\text{getting a sum of 7 given that one of the numbers is odd.})=\frac{5}{36}$

(15)Probability of getting a sum of eight given that both numbers are even numbers.

The possible outcomes are: (2, 6), (4, 4), (6, 2)

$P(\text{getting a sum of eight given that both numbers are even numbers.})=\frac{3}{36}\\=\frac{1}{12}$

16.Probability of getting two numbers with a sum of 14.

$P(\text{getting two numbers with a sum of 14.})=\frac{0}{36}=0$

4 0

4 years ago