What is the domain of (1,2),(2,6),(3,10),(2,14),(1,0)

Mathematics

1 answer:

Bas_tet [7]3 years ago

5 0

The set of x ordinates or {1,2,3}

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Which of the following are solutions to the quadratic equation? Check all that

Troyanec [42]

Answer:

x = 1/3 or -2

Step-by-step explanation:

2x^2 + 7x - 2 = - x^2 + 2x

3x^2 + 5x - 2 = 0 (use factoring)

(3x − 1)(x + 2) = 0

3x - 1 = 0 OR x + 2 = 0

3x = 1 x = - 2

x = 1/3

Can you mark as brainliest? Hope it helps by the way!

3 0

3 years ago

Read 2 more answers

Find sin(a+b) if tan(a)=7/24 where a is in the third quadrant

ludmilkaskok [199]

The complete question in the attached figure

we have that
tan a=7/24 a----> III quadrant
cos b=-12/13 b----> II quadrant
sin (a+b)=?

we know that
sin(a + b) = sin(a)cos(b) + cos(a)sin(b<span>)
</span>
step 1
find sin b
sin²b+cos²b=1------> sin²b=1-cos²b----> 1-(144/169)---> 25/169
sin b=5/13------> is positive because b belong to the II quadrant

step 2
Find sin a and cos a
tan a=7/24
tan a=sin a /cos a-------> sin a=tan a*cos a-----> sin a=(7/24)*cos a
sin a=(7/24)*cos a------> sin²a=(49/576)*cos²a-----> equation 1
sin²a=1-cos²a------> equation 2
equals 1 and 2
(49/576)*cos²a=1-cos²a---> cos²a*[1+(49/576)]=1----> cos²a*[625/576]=1
cos²a=576/625------> cos a=-24/25----> is negative because a belong to III quadrant
cos a=-24/25
sin²a=1-cos²a-----> 1-(576/625)----> sin²a=49/625
sin a=-7/25-----> is negative because a belong to III quadrant

step 3
find sin (a+b)
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin a=-7/25
cos a=-24/25
sin b=5/13
cos b=-12/13
so
sin (a+b)=[-7/25]*[-12/13]+[-24/25]*[5/13]----> [84/325]+[-120/325]
sin (a+b)=-36/325

the answer is
sin (a+b)=-36/325