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bixtya [17]
3 years ago
10

Find the exact value of sin 165º.

Mathematics
2 answers:
ra1l [238]3 years ago
8 0

Answer:0.2588

Step-by-step explanation:

lisabon 2012 [21]3 years ago
4 0

Answer:

\frac{1}{4} ( \sqrt{6} - \sqrt{2} )

Step-by-step explanation:

Using the sine addition formula

sin(a + b) = sinacosb + cosasinb

and the exact values

sin45° = cos45° = \frac{\sqrt{2} }{2} , sin60° = \frac{\sqrt{3} }{2} , cos60° = \frac{1}{2}

Given

sin165°

Note that 165° = (120 + 45)°, thus

= sin(120 + 45)°

= sin120°cos45° + cos120°sin45°

= sin60°cos45° - cos60°sin45°

= ( \frac{\sqrt{3} }{2} × \frac{\sqrt{2} }{2} ) - ( \frac{1}{2} × \frac{\sqrt{2} }{2} )

= \frac{\sqrt{6} }{4} - \frac{\sqrt{2} }{4}

= \frac{1}{4} ( \sqrt{6} - \sqrt{2} )

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uranmaximum [27]

Answer:

x=2 , y = 1

Step-by-step explanation:

x + 2(2x-3) = 4

x+4x-6=4

5x-6=4

5x= 10

x=10/5

x=2

subs x=2 into x +2y=4

2+2y=4

2y=2

y=2/2

y=1

6 0
3 years ago
Read 2 more answers
Someone help please haha​
harkovskaia [24]

Answer:

{-8, -7, 0, 6, 9}

Step-by-step explanation:

1. The range of a relation is the set of its possible output values, also known as the y-values of a function.

2. Let's find the y-coordinate of each point.

  • 9, 6, 0, -7, -8

3. Now, let's order them (from least to greatest) to get the range.

  • -8, -7, 0, 6, 9
  • {-8, -7, 0, 6, 9}

Therefore, the range of this relation is {-8, -7, 0, 6, 9}.

8 0
3 years ago
Please help me to solve this equation<br><br><img src="https://tex.z-dn.net/?f=ax%20-%20by%20%3E%20c" id="TexFormula1" title="ax
slega [8]

Given: ax-by>c

Subtract ax on both sides: -by>-ax+c

Divide both sides by -b: y<(-ax+c)/-b

Simplify: y<(ax-c)/b

your answer: y<(ax-c)/b


8 0
3 years ago
The digit 1 in which number represents a value of 1 tenth
sattari [20]

Answer:

0.1

Step-by-step explanation:

6 0
3 years ago
Christian is rewriting an expression of the form y = ax2 + bx + c in the form y = a(x – h)2 + k. Which of the following must be
notsponge [240]
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"k and c have the same value" -- k and c do not have the same value. "k" is the y-value of the vertex and c is the constant in your quadratic equation, and the constant is not necessarily the y-value.

"the value of a remains the same" -- this is true. the a's in your equations are the same values, because the a-value is the coefficient of the x-variable in both equations. y = a(x - h)^2 and y = ax^2 -- both of these have a applying to your x-variables.

"h is equal to one half -b" -- this isn't true. the formula for calculating the x value of the vertex (h is the x-value of the vertex) is h = (-b/2a). -b/2a is not the same as one half -b because this answer choice doesn't involve the a-value.
3 0
2 years ago
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