Step-by-step explanation:
All the steps are in the pictures. If you have any questions let me know and I'll explain in simple terms :)
Answer: hello some part of your question is missing
Let v=〈−2,5〉 in R^2,and let y=〈0,3,−2〉 in R^3.
Find a unit vector u in R^2 such that u is perpendicular to v. How many such vectors are there?
answer:
One(1) unit vector ( < 5/√29, 2 /√29 > ) perpendicular to 〈−2,5〉
Step-by-step explanation:
let
u = < x , y > ∈/R^2 be perpendicular to v = < -2, 5 > ------ ( 1 )
hence :
-2x + 5y = 0
-2x = -5y
x = 5/2 y
back to equation 1
u = < 5/2y, y >
∴ || u || = y/2 √29
∧
u = < 5 /2 y * 2 / y√29 , y*2 / y√29 >
= < 5/√29, 2 /√29 > ( unit vector perpendicular to < -2, 5 > )
It will take her 30 minutes to type it because if you take 2100 and divide it by 350 you’ll get 6,so that how many times it fix’s into it,then do 6x5 then you get your answer
Answer:
D. 70°
Step-by-step explanation:
Given:
m<DGB = 35°
m<CGF = 75°
Required:
m<AGE
SOLUTION:
m<EGD = m<CGF (vertical angles are congruent)
m<EGD = 75°
m<AGE + m<EGD + m<DGB = 180° (angles on a straight line)
m<AGE + 75° + 35° = 180° (substitution)
m<AGE + 110° = 180°
Subtract 110 from each side of the equation
m<AGE = 180° - 110°
m<AGE = 70°
Answer:
i think 6x
Step-by-step explanation:
cz we have a variable