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Naya [18.7K]
2 years ago
5

Can someone help me???

Mathematics
2 answers:
cestrela7 [59]2 years ago
7 0

Answer:

Cos (5π/3)

Cos (π/6)I think this one

Lemur [1.5K]2 years ago
6 0

Answer:

Below in bold.

Step-by-step explanation:

The sine is positive in the first and second quadrants and negative in other 2.

The cosine is positive in the first and 4th quadrants and negative in other 2.

Note that sin(pi/6) = 1/2 and cos(pi/3 = 1/2).

sin(7pi/6) = 1/2  

cos(5pi/3) = 1/2

sin(5pi/6) = 1/2

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Answers fast please. 1.) y=x−63x+2y=8 Use the substitution method. A.) (4, −2) B.) (14, 8) C.) (0, −6) D.) (3, −3) 2.) What is t
GrogVix [38]

<u>QUESTION 1</u>

The given system of equation is

y=x-6...eqn1

and


3x+2y=8...eqn2


Let us substitute equation (1) into equation (2) to get,

3x+2(x-6)=8


We expand the bracket to get,

3x+2x-12=8


We simplify to get.

5x-12=8


We group like terms to get

5x=8+12

\Rightarrow 5x=20

\Rightarrow x=4


We now substitute x=4 in to equation (1) to obtain,


y=4-6=-2


The correct answer is option A.


<u>QUESTION 2</u>

The given system of equations is

y-x=9...eqn1


and

10+2x=-2y...eqn2

We make y the subject in equation (2) to get,

y=-x-5..eqn3


We put equation (3) into equation (1) to obtain,


-x-5-x=9


We group like terms to get,

-x-x=9+5


This implies that,

-2x=14


We divide through by -2 to get,

x=-7


Hence the x-coordinate is -7



<u>QUESTION 3</u>

The given system is

y=3x-1..eqn1


and


x-y=-9...eqn2

We make x the subject in equation (2) to get,

x=y-9...eqn3


We put equation (3) into equation (1) to obtain,


y=3(y-9)-1


We expand the bracket to get,


y=3y-27-1


Group like terms to get,


y-3y=-27-1


We simplify to get;

-2y=-28


This implies that,

y=14


Therefore the y-coordinate is 14.

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