The vector ab has a magnitude of 20 units and is parallel to the
vector 4i + 3j. Hence, The vector AB is 16i + 12j.
<h3>How to find the vector?</h3>
If we have given a vector v of initial point A and terminal point B
v = ai + bj
then the components form as;
AB = xi + yj
Here, xi and yj are the components of the vector.
Given;
The vector ab has a magnitude of 20 units and is parallel to the
vector 4i + 3j.
magnitude
Unit vector in direction of resultant = (4i + 3j) / 5
Vector of magnitude 20 unit in direction of the resultant
= 20 x (4i + 3j) / 5
= 4 x (4i + 3j)
= 16i + 12j
Hence, The vector AB is 16i + 12j.
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Answer: 23
Step-by-step explanation:
AB + BC = AC
AC = 11 + 12 = <u>23</u>
Answer:
S(-2, -3)
Step-by-step explanation:
Find the diagram attached below,=. Frim the diagram, the coordinate of R and T are (-5, 3) and (-1, -5) respectively. If the ratio of RS to ST is 3:1, the coordinate of S can be gotten using the midpoint segment formula as shown;
S(X, Y) = {(ax1+bx2/a+b), (ay1+by1/a+b)} where;
x1 = -5, y1 = 3, x2 = -1, y2 = -5, a = 3 and b =1
Substitute the values into the formula;
X = ax2+bx1/a+b
X = 3(-1)+1(-5)/3+1
X = -3-5/4
X = -8/4
X = -2
Similarly;
Y = ay2+by1/a+b
Y = 3(-5)+1(3)/3+1
Y = -15+3/4
Y = -12/4
Y = -3
Hence the coordinate of the point (X, Y) is (-2, -3)
Given: f(x)=(2x-2)/4
find f^-1(3)?
we need to find the inverse of f(x), so
x=(2y-2)/4
2y-2=4x
y-1=2x
y=2x+1
so, then f^-1(x)=2x+1
f^-1(3)=2(3)+1
=6+1
=7
so, the answer is 7
