Answer:(1,1)
Step-by-step explanation:
Its asking where both linear and curve lines are being hit,
(fxg)(0)= 1,1
9.77 to the nearest tenths
first, find the number that is in the tenths place....it is 7....now look at the number directly to the right of it....if that number is 5 or greater, u would round that 7 up to an 8...but if that number is 4 or below, ur 7 would stay the same.
So the number directly to the right of 7 is 7...and since it is greater then 5, u have to round the 7 in the tenths place up to 8.
solution is : 9.8
Separate the vectors into their <em>x</em>- and <em>y</em>-components. Let <em>u</em> be the vector on the right and <em>v</em> the vector on the left, so that
<em>u</em> = 4 cos(45°) <em>x</em> + 4 sin(45°) <em>y</em>
<em>v</em> = 2 cos(135°) <em>x</em> + 2 sin(135°) <em>y</em>
where <em>x</em> and <em>y</em> denote the unit vectors in the <em>x</em> and <em>y</em> directions.
Then the sum is
<em>u</em> + <em>v</em> = (4 cos(45°) + 2 cos(135°)) <em>x</em> + (4 sin(45°) + 2 sin(135°)) <em>y</em>
and its magnitude is
||<em>u</em> + <em>v</em>|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)
… = √(16 cos²(45°) + 16 cos(45°) cos(135°) + 4 cos²(135°) + 16 sin²(45°) + 16 sin(45°) sin(135°) + 4 sin²(135°))
… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))
… = √(16 + 16 cos(135° - 45°) + 4)
… = √(20 + 16 cos(90°))
… = √20 = 2√5
Given:
To verify:
for the given values.
Solution:
We have,
We need to verify .
Taking left hand side, we get
Taking LCM, we get
Taking right hand side, we get
Taking LCM, we get
Now,
Hence proved.
