Which numbers are integers 4 -3 42/6 -5.6 5.423 -54 98 54/6

Please take a look at the picture. Please write your answer with an explanation.

Ahat [919]

Step-by-step explanation:

what answer is needed ? the result of the calculation ?

I assume this is the case.

remember that regarding the sequence of operations brackets are first, then exponents, then "dot" operations (× and ÷), and only at the very least come the "dash" operations (+ and -).

and that a combination of - + or + - results in -.

14 + 16 ÷ 2³ - (12 + 3²) ÷ 7 + 2 × (-5) =

14 + 16 ÷ 2³ - (12 + 9) ÷ 7 - 2 × 5 =

14 + 16 ÷ 2³ - 21 ÷ 7 - 2 × 5 =

14 + 16 ÷ 8 - 21 ÷ 7 - 2 × 5 =

14 + 2 - 3 - 10 = 3

7 0

2 years ago

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)

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