Answer:
Step-by-step explanation:
1. Eat 2. Leave 6.bye
y = 3
The opposite sides of a rectangle are equal thus
2y + 4 = 10
subtract 4 from both sides
2y = 10 - 4 = 6
divide both sides by 2
y = 
= 3
We know that
if cos x is positive
and
sin x is negative
so
the angle x belong to the IV quadrant
cos x=5/13
we know that
sin²x+cos²x=1-------> sin²x=1-cos²x------> 1-(5/13)²---> 144/169
sin x=√(144/169)-------> sin x=12/13
but remember that x is on the IV quadrant
so
sin x=-12/13
Part A) <span>cos (x/2)
cos (x/2)=(+/-)</span>√[(1+cos x)/2]
cos (x/2)=(+/-)√[(1+5/13)/2]
cos (x/2)=(+/-)√[(18/13)/2]
cos (x/2)=(+/-)√[36/13]
cos (x/2)=(+/-)6/√13-------> cos (x/2)=(+/-)6√13/13
the angle (x/2) belong to the II quadrant
so
cos (x/2)=-6√√13/13
the answer Part A) is
cos (x/2)=-6√√13/13
Part B) sin (2x)
sin (2x)=2*sin x* cos x------> 2*[-12/13]*[5/13]----> -120/169
the answer Part B) is
sin(2x)=-120/169
Answer:
Step-by-step explanation:
You need to use distributive property and multiply each term of the binomial "3x+7" (that is 3x, and 7) by each term of the trinomial "
" taking care of their appropriate signs, and then combine like terms until there is no more than one term for each power of "x". It is also customary to order the final terms in descending order of power of "x".
Let's start by multiplying as indicated above:
![(2x^2+5x-3)*(3x+7)=\\=2x^2\,*\,3x+2x^2\,*\,7\,+5x*3x+5x*7-3*3x-3*7=\\=6x^3+14x^2+15x^2+35x-9x-21=\\=6x^3+[14x^2+15x^2]+[35x-9x]-21=\\=6x^3+29x^2+26x-21](https://tex.z-dn.net/?f=%282x%5E2%2B5x-3%29%2A%283x%2B7%29%3D%5C%5C%3D2x%5E2%5C%2C%2A%5C%2C3x%2B2x%5E2%5C%2C%2A%5C%2C7%5C%2C%2B5x%2A3x%2B5x%2A7-3%2A3x-3%2A7%3D%5C%5C%3D6x%5E3%2B14x%5E2%2B15x%5E2%2B35x-9x-21%3D%5C%5C%3D6x%5E3%2B%5B14x%5E2%2B15x%5E2%5D%2B%5B35x-9x%5D-21%3D%5C%5C%3D6x%5E3%2B29x%5E2%2B26x-21)
use a calculator it's way easier but here is the answer
