Take the cross product, then normalize the result.
This has norm
and so a unit vector orthogonal to both given vectors is
An equally correct answer would be the negative of this vector, since
.
Answer:
First
Step-by-step explanation:
The first person in the line will always have an odd position, not an even one. They will be the only one left.
Answer:
58 units squared
Step-by-step explanation:
We want to find the area of the square. To do so, we need to find the hypotenuse of the right triangle because this coincides with the side length of the square.
We use the Pythagorean Theorem, which states that for a right triangle with legs a and b and hypotenuse c:
a^2 + b^2 = c^2
Here, a = 7 and b = 3, so:
7^2 + 3^2 = c^2
c^2 = 49 + 9 = 58
Now, the area of a square is: A = s^2, where s is the side length. Well, c is the side length, and we've already found what c^2 is (it's 58), so that means the area of the square is 58 units squared.
Thus, the answer is 58 units squared.
Answer:
Step-by-step explanation:
Ok, so first substitute x for 4 in
"g(x) = 5x + 1", and x for 3 in
"k(x) = 2/x + 2x". Now you got:
g(4) = 5(4) + 1
k(3) = 2/(3) + 2(3)
Now you can solve each individually.
g(4) = 5 × 4 = 20
20 + 1 = 21
g(4) = 21
k(3) = 2 × 3 = 6
6 + 2/3 = 6 2/3
k(3) = 6 2/3
g(4) + k(3) = 21 + 6 2/3 = <u>27 2/3</u>
Hope this helps :)
Answer:
Step-by-step explanation:
<u>For old circular garden:</u>
take the radius as r.
then use the formula to find area of circle: πr² ......this is old garden area.
<u>For new enlarged garden:</u>
the radius is twice the old radius so, radius = 2 * r = 2r ......enlarged radius
now find area for this new garden: π(2r)² → 4πr²
In common fractions: (old garden)/(new garden)
: ( πr² ) / ( 4πr² )
: 1/4
