The vector i=<1,0> and j=<0,1> so the i+j=<1+0,0+1>=<1,1>. The length of this vector is easy: |i+j|=<span>2–√</span>
to make the vector i+j=<1,1> a unit vector we rescale it by it's
length (i.e. divide i+j by its length) , v=(i+j)/(|i+j|)
thus we have v=<span>1/<span>2–√</span><1,1></span> or <span><1/<span>2–√</span>,1/<span>2–√</span>></span>
If you check the length of this vector v, you see it indeed does have
length =1. It is parallel to the vector i+j because it's components are
proportional to the components of i+j=<1,1>.
Answer:
option A)
(40, 96)
Step-by-step explanation:
Given that,
The coordinates of point K and J are
K(160,120)
J(-40,80)
x1 = 160
x2 = -40
y1 = 120
y2 = 80
P is (3/5) the line of the line segment from K to J
So, KP = (3/5) KJ and JP = (2/5) KJ
OR we will divide the length of KJ with the ratio 3 : 2 from K
m : n
3 : 2
m = 3
n = 2
by using this formula and putting values in it
xp = (m/m+n)(x2-x1) + x1
yp = (m/m+n)(y2-y1) + y1
xp = (3/3+2) (-40-160) + 160
yp = (3/3+2) (80-120) + 120
xp = 40
yp = 96
Answer:
The answer is Median.
Step-by-step explanation:
Problem One
Test it.
a^2 + b^2 = c^2
a = 8
b = 12 What will c calculate to be.
8^2 + 12^12 = c^2
c^2 = 64 + 144
c^2 = 208
If 18 is to be the hypotenuse of the triangle c^ must equal 324.
208 is nowhere's near that amount.
Mario is right.
Question Two
The only way Hank will be right is if 26 is the hypotenuse. If that is so, the third side is 24. This is how you do it.
a^2 + b^2 = c^2
c = 26
a = 10
10^2 + b^2 = 26^2
100 + b^2 = 676 Subtract 100
b^2 = 676 - 100
b^2 = 576
b = sqrt(576)
b = 24
If 26 is not the hypotenuse, the right triangle cannot be drawn.
