For every negative power you divide by 10
Trapezoidal is involving averageing the heights
the 4 intervals are
[0,4] and [4,7.2] and [7.2,8.6] and [8.6,9]
the area of each trapezoid is (v(t1)+v(t2))/2 times width
for the first interval
the average between 0 and 0.4 is 0.2
the width is 4
4(0.2)=0.8
2nd
average between 0.4 and 1 is 0.7
width is 3.2
3.2 times 0.7=2.24
3rd
average betwen 1.0 and 1.5 is 1.25
width is 1.4
1.4 times 1.25=1.75
4th
average betwen 1.5 and 2 is 1.75
width is 0.4
0.4 times 1.74=0.7
add them all up
0.8+2.24+1.75+0.7=5.49
5.49
t=time
v(t)=speed
so the area under the curve is distance
covered 5.49 meters
Answer:x=3
Step-by-step explanation:
D( x )
x+2 = 0
x = 0
x+2 = 0
x+2 = 0
x+2 = 0 // - 2
x = -2
x = 0
x = 0
x in (-oo:-2) U (-2:0) U (0:+oo)
(9*x-7)/(x+2)+15/x = 9 // - 9
(9*x-7)/(x+2)+15/x-9 = 0
(x*(9*x-7))/(x*(x+2))+(15*(x+2))/(x*(x+2))+(-9*x*(x+2))/(x*(x+2)) = 0
x*(9*x-7)+15*(x+2)-9*x*(x+2) = 0
9*x^2-9*x^2+8*x-18*x+30 = 0
30-10*x = 0
(30-10*x)/(x*(x+2)) = 0
(30-10*x)/(x*(x+2)) = 0 // * x*(x+2)
30-10*x = 0
30-10*x = 0 // - 30
-10*x = -30 // : -10
x = -30/(-10)
x = 3
x = 3
A x = 0
using the law of exponents = 1
for (6² )^ x = 1 then x = 0
B note that = 1 ⇒ x = 1
2 → 2^8 × 3^(-5) × 1^(-2) × 3^(-8) × 2^(-12) × 2^(28)
= 2^(8 -12 + 28) × 1 × 3^(- 5 - 8)
= 2^24 × 3^(- 13) = 2^(24)/3^(13) = 10.523 ( 3 dec. places)
Answer:
After 7 years, the value would be $9,046.
After 13 years, the value would be $3,660
Step-by-step explanation: