Answer:
solo suma todos los valores x+4(3)+(x+x)
en el segundo x+5(2)+x+3(2)
en el tercero x-3+(2x+5)+(2x+5)
y en el cuarto x(3)+(x+4)
2.4+3.06=5.46
5.46+.75=6.21
The answer is 6.21
11. Take the logarithm, base X.
a² - ab = 10
Since we are given that b = a - 2, we can substitute for b to get
a² - a(a -2) = 10
a² - a² + 2a = 10 . . . . . eliminate parentheses
a = 5 . . . . . . . . . . . . . . simplify, divide by 2
_____
Check: a² = 5² = 25
ab = 5·3 = 15
X^25/X^15 = X^10 . . . . . the answer works
Answer:
1. 1 point
2. The x-coordinate of the solution = 2/17
3. The y-coordinate of the solution = -16/17
Step-by-step explanation:
Given that the equation is of the form;
y = -2³×x and y = 9·x - 2, we have;
y = -8·x and y = 9·x - 2
1. Given that the two lines are straight lines, the number of points of intersection is one.
2. The x-coordinate of the solution
To find a solution to the system of equations, we equate both expression of the functions and solve for the independent variable x as follows;
-8·x = 9·x - 2
-8·x - 9·x= - 2
-17·x = -2
x = 2/17
The x-coordinate of the solution = 2/17
3. The y-coordinate of the solution
y = 9·x - 2 = 9×2/17 - 2 = -16/17
y = -16/17
The y-coordinate of the solution = -16/17.
We have to assume that the speed before being stuck was sufficient to get to the destination on time had there been no delay. Call that speed "s" in km/h.
Since 200 km is "halfway", the total distance must be 400 km.
time = distance / speed
total time = (time for first half) + (delay) + (time for second half)
400/s = 200/s + 1 + 200/(s+10) . . . .times are in hours, distances in km
200/s = 1 + 200/(s+10) . . . . . . . . . . subtract 200/s
200(s+10) = s(s+10) +200s . . . . . . .multiply by s(s+10)
0 = s² +10s - 2000 . . . . . . . . . . . . . .subtract the left side
(s+50)(s-40) = 0
Solutions are s = -50, s = 40
The speed of the bus before the traffic holdup was 40 km/h.
