Answer:
(1 , 1)
Step-by-step explanation:
Given in the question two equations
Equation 1
y = x² – x + 1
Equation 2
y = x
Equate both equations
x² – x + 1 = x
0 = x² – x + 1 - x
x² – 2x + 1 = 0
x² - x - x + 1 = 0
x(x - 1) -1(x - 1) = 0
(x - 1)(x - 1) = 0
(x-1)² = 0
x - 1 = 0
x = 1
Plug x = 1 in first equation
y = 1² – 1 + 1
y = 1
Answer:
155°
Step-by-step explanation:
∠EYV is half the difference of arcs EV and EH
(1/2)(EV -EH) = ∠EYV
(1/2)(EV -85°) = 35° . . . . fill in the given values
EV -85° = 70° . . . . . . . . .multiply by 2
EV = 155° . . . . . . . . . . . . add 85°
(0,346)(2,344.8)
slope = (344.8 - 346) / (2 - 0) = -1.2 / 2 = -0.6
y = mx + b
slope(m) = -0.6
use either of ur points (0,346)...x = 0 and y = 346
now we sub and find b, the y int
346 = -0.6(0) + b
346 = b
so ur equation is : y = -0.6x + 346
after 4 weeks....x = 4
y = -0.6(4) + 346
y = -2.4 + 346
y = 343.6 <=== after 4 weeks it will be 343.6
The average speed of Joshua during that time is 2500 m/h.
Explanation:
It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.
The total time taken by Joshua from 5:15 pm to 8:09 pm is
Dividing we get,
Adding, we have,
Thus, the total time taken by Joshua is 
To determine the average speed we use the formula,
where 
and 
Hence, substituting the values we have,
Dividing, we get,
Thus, the average speed of Joshua during that time is 2500 m/h.
