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Reptile [31]
3 years ago
13

Find sin(a+B)

Mathematics
1 answer:
diamong [38]3 years ago
6 0

Answer:

Cos A=5/13

we have

Cos² A=1-Sin ²A

25/169=1-Sin²A

sin²A=1-25/169

sin²A=144/169

Sin A=\sqrt{144/169}=12/13

again

Tan B=4/3

P/b=4/3

p=4

b=3

h=\sqrt{3²+4²}=5

Now

Sin B=p/h=4/5

in IV quadrant sin angle is negative so

Sin B=-4/5

CosB=b/h=3/5

Now

<u>S</u><u>i</u><u>n</u><u>(</u><u>A</u><u>+</u><u>B</u><u>)</u><u>:</u><u>s</u><u>i</u><u>n</u><u>A</u><u>c</u><u>o</u><u>s</u><u>B</u><u>+</u><u>C</u><u>o</u><u>s</u><u>A</u><u>s</u><u>i</u><u>n</u><u>B</u>

<u>n</u><u>o</u><u>w</u><u> </u>

<u>substitute</u><u> </u><u>value</u>

<u>Sin(A+B):</u>12/13*3/5+5/13*(-4/5)=36/65-4/13

<u>=</u><u>1</u><u>6</u><u>/</u><u>6</u><u>5</u><u> </u><u>i</u><u>s</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u>

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Answer:

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6 0
3 years ago
Please show your work!!!!
viva [34]

Answer:

4√34

Step-by-step explanation:

Let the unknown side be y

y^2 = 20^2 + 12^2

y^2 = 400 + 144

y^2 = 544

Take the square root of both side

y = √544

y = √(16x34)

y = √16 x √34

y = 4√34

6 0
3 years ago
1. Write an equation in point-slope form of the line that that passes through the point (4,-5) and has a
Aleks04 [339]

Answer:

y + 5 = (3/2)(x - 4)

Step-by-step explanation:

The point-slope formula is y - k = m(x - h).  We get h and k from the given point:  h = 4 and k = -5, and m from the given slope:  3/2.  Then we have:

y + 5 = (3/2)(x - 4)

7 0
3 years ago
What is 17- 6 11/12<br> Can someone please solve in steps?
Airida [17]

17- 6 11/12

borrow from the 17 and use a common denominator

16 12/12 - 6 11/12

10 1/12

6 0
3 years ago
You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob
Ilia_Sergeevich [38]

Answer:

a) No

b) 42%

c) 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c)   P(win first game)  = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game)  × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

P(X = 0)  =  P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game)  × P(win second game)] + [ P(win first game)  × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value  \mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66

f) Variance \sigma^2=\Sigma(x-\mu^2)p(x)= (0-0.66)^2*0.42+ (1-0.66)^2*0.5+ (2-0.66)^2*0.08=0.3844

Standard deviation \sigma=\sqrt{variance} = \sqrt{0.3844}=0.62

8 0
3 years ago
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