Answer:
16 or 14 depending on that 3 after the first x
Step-by-step explanation:
(2)^3+5(2)-2
8+10-2
18-12=16
OR
(2)3+5(2)-2
6+10-2
16-2=14
15 mi/h= 15mi/60 min= 1/4 (mi/min)
x is original speed (mi/min)
t is original time
x*t =260
(x+1/4)=5x/4 is new speed
t-20 - new time
(x+1/4)(t-20)=260
xt = (x+1/4)(t-20)
xt=xt + t/4 - 20x - 5
t/4=20x+5
t=80x +20
We can substitute t=80x +20 into equation x*t =260
x*(80x+20)=260
80x²+20x-260=0
4x²+x-13=0
D=b² - 4ac = 1+4*4*13= 209
x=(-1+/-√209)/2*4
x=1.68 mi/min
1.68mi/min * 60 min/1h≈100 mi/h
Check
260/100=2.6h time with old speed
260/(115)=2.26 h time with new speed
2.6-2.26 = 0.34 h difference between old and new time
0.34h*60min/1h≈20 min difference between old and new time in minutes
Yes I can but what’s the question???
perp to x-3y=2 thru (2,4)
For perpendicular we swap the coefficients on x and y, negating one
3x + y = some constant
We get the constant in the obvious way from the point
3x + y = 3(2) + 4 = 6 + 4 = 10
Answer: 3x + y = 10
