Answer:
(x, y) = (-3, 2)
Step-by-step explanation:
Put the given value of y into the equation and solve for x.
... 7·2 +3x = 5
... 3x = -9 . . . . . . subtract 14
... x = -3 . . . . . . . divide by the x-coefficient
The solution (x, y) is (-3, 2).
3w+w-2
4w-2
GCF is 2
2(2w-1)
Answer:
A) (1 s, 2.3 s)
B) (-4 m/s², 3.8 m/s²)
Step-by-step explanation:
The car's position which is the distance is given by the equation;
s(t) = t³ - 5t² + 7t
A) Velocity is the first derivative of the distance. Thus;
v(t) = ds/dt = 3t² - 10t + 7
At v = 0, we have;
3t² - 10t + 7 = 0
Using quadratic formula, we have;
t = 1 and t = 2.3
Thus, time at velocity of 0 is t = (1 s, 2.3 s)
B) acceleration is the derivative of the velocity. Thus;
a(t) = dV/dt = 6t - 10
At velocity of 0, we got t = 1 and t = 2.3
Thus;
a(1) = 6(1) - 10 = -4 m/s²
a(2.3) = 6(2.3) - 10 = 3.8 m/s
Thus, a(t) at v = 0 gives; (-4 m/s², 3.8 m/s²)
a) For x = 27:
z = 27-28/2 = -0.5
For x = 31:
z = 31-38/2 = 1.5
From the normal distribution table, P(27 < x < 31) = P(-0.5 < z < 1.5) = P(z < 1.5) - P(z < -0.5) = 0.9332 - 0.3085 = 62.47%
b) For x > 30.2:
z = 30.2-28/2 = 1.1
From the normal distribution table, P(x > 30.2) = P(z > 1.1) = 1 - P(z > 1.1) = 1 - 0.8643 = 13.57%
Multiply the denominator by the number in front of it and then add it to the numerator. (6•2=12+5=17). The denominator stays the same. Your answer is 17/6. Hope this helped!
