Answer:
The answer is
<h2>( 6 , 4 )</h2>
Step-by-step explanation:
The midpoint M of two endpoints of a line segment can be found by using the formula
where
(x1 , y1) and (x2 , y2) are the points
From the question the points are
(3, 2) and (9, 6)
The midpoint is
We have the final answer as
<h3>( 6 , 4 )</h3>
Hope this helps you
Answer:
x = 18 (Smaller number)
y = 35 (Larger number)
Step-by-step explanation:
To solve, begin by setting up a system of equations.
'x' being the smaller number, while
'y' is the larger number.
We can construct the equations:
x + y = 53
3x = y + 19
Rearrange the second equation to equal 'y':
3x - 19 = y
Plug this value of 'y' into the first equation:
x + 3x - 19 = 53
Combine like terms and simplify:
4x - 19 = 53
4x = 72
x = 18
Plug in this value for 'x' into an equation to solve for 'y':
18 + y = 53
y = 35.
Answer:
Step-by-step explanation:
First, simplify each term:
Then given expression is equivalent to
![\cos ^3\alpha+(-\sin \alpha)^3-(-\sin \alpha)+(-\cos \alpha)\\ \\=\cos ^3\alpha-\sin^3 \alpha+\sin \alpha-\cos \alpha\\ \\=(\cos\alpha-\sin\alpha)(\cos^2\alpha+\cos\alpha\sin\alpha+\sin^2\alpha)-(\cos\alpha-\sin\alpha)\\ \\=(\cos\alpha-\sin\alpha)(1+\cos\alpha\sin\alpha-1)\ \ [\cos^2\alpha+\sin^2\alpha=1]\\ \\=\cos\alpha\sin\alpha(\cos\alpha-\sin\alpha)](https://tex.z-dn.net/?f=%5Ccos%20%5E3%5Calpha%2B%28-%5Csin%20%5Calpha%29%5E3-%28-%5Csin%20%5Calpha%29%2B%28-%5Ccos%20%5Calpha%29%5C%5C%20%5C%5C%3D%5Ccos%20%5E3%5Calpha-%5Csin%5E3%20%5Calpha%2B%5Csin%20%5Calpha-%5Ccos%20%5Calpha%5C%5C%20%5C%5C%3D%28%5Ccos%5Calpha-%5Csin%5Calpha%29%28%5Ccos%5E2%5Calpha%2B%5Ccos%5Calpha%5Csin%5Calpha%2B%5Csin%5E2%5Calpha%29-%28%5Ccos%5Calpha-%5Csin%5Calpha%29%5C%5C%20%5C%5C%3D%28%5Ccos%5Calpha-%5Csin%5Calpha%29%281%2B%5Ccos%5Calpha%5Csin%5Calpha-1%29%5C%20%5C%20%5B%5Ccos%5E2%5Calpha%2B%5Csin%5E2%5Calpha%3D1%5D%5C%5C%20%5C%5C%3D%5Ccos%5Calpha%5Csin%5Calpha%28%5Ccos%5Calpha-%5Csin%5Calpha%29)
Answer:
Yes.
Step-by-step explanation:
Just like normal algebra, you factor our the common factor, in this case, 5.
Thus,
