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docker41 [41]
3 years ago
15

The point (2, 3) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ

?
Mathematics
1 answer:
Ganezh [65]3 years ago
8 0
(2,3) so x=2, y=3 and h=(2^2+3^2)^(1/2)=√13

sina=3/√13, cosa=2/√13, tana=3/2
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Find the equation of a line passing through the point (6,−5) and parallel to the line y = 7x + 1. answers : y = 7x − 47 y = −17x
Elan Coil [88]

Answer:

<h2>y = 7x-47</h2>

Step-by-step explanation:

y = 7x + 1.\\(6,-5)\\m= 7\\x1= 6\\y1 = -5\\y-y1=m(x-x1)\\y-(-5) = 7(x-6)\\y+5=7x-42\\y=7x-42-5\\y = 7x-47

3 0
3 years ago
7n +9 = -1+6(n+4)<br>can someone walk me through this please?​
Ira Lisetskai [31]

Answer:

n = 14

Step-by-step explanation:

7n +9 = -1+6(n+4)

Distribute

7n+9 = -1 +6n+24

Combine like terms

7n +9 = 6n +23

Subtract 6n from each side

7n -6n+9 = 6n-6n +23

n+9 = 23

Subtract 9 from each side

n+9-9=23-9

n = 14

6 0
3 years ago
Hey could someone help me with this question immediately!! i’ll mark brainiest if it’s correct.
Bezzdna [24]

the thousands placeeee

5 0
3 years ago
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Rewrite the expression below.<br><br> -2a( a + b - 5)+ 3 (-5a + 2b) +b (6a + b - 8)
Paul [167]

Answer:

b^2 + 4 a b - 2 b - 2 a^2 + -5 a

Step-by-step explanation:

Simplify the following:

-2 a (a + b - 5) + 3 (2 b - 5 a) + b (6 a + b - 8)

-2 a (-5 + a + b) = 10 a - 2 a^2 - 2 a b:

10 a - 2 a^2 - 2 a b + 3 (2 b - 5 a) + b (6 a + b - 8)

3 (2 b - 5 a) = 6 b - 15 a:

10 a - 2 a^2 - 2 a b + 6 b - 15 a + b (6 a + b - 8)

b (-8 + 6 a + b) = -8 b + 6 a b + b^2:

10 a - 2 a^2 - 2 a b - 15 a + 6 b + -8 b + 6 a b + b^2

Grouping like terms, 10 a - 2 a^2 - 2 a b - 15 a + 6 b - 8 b + 6 a b + b^2 = b^2 + (6 a b - 2 a b) + (6 b - 8 b) - 2 a^2 + (10 a - 15 a):

b^2 + (6 a b - 2 a b) + (6 b - 8 b) - 2 a^2 + (10 a - 15 a)

a b 6 + a b (-2) = 4 a b:

b^2 + 4 a b + (6 b - 8 b) - 2 a^2 + (10 a - 15 a)

6 b - 8 b = -2 b:

b^2 + 4 a b + -2 b - 2 a^2 + (10 a - 15 a)

10 a - 15 a = -5 a:

Answer:  b^2 + 4 a b - 2 b - 2 a^2 + -5 a

8 0
3 years ago
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tatiyna
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