Answer:
Step-by-step explanation:
![\sqrt[3]{2x^5y^7} * \sqrt[3]{4x^4y^2} \\\\2x^3y^3](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%5E5y%5E7%7D%20%2A%20%5Csqrt%5B3%5D%7B4x%5E4y%5E2%7D%20%5C%5C%5C%5C2x%5E3y%5E3)
Answer:
x = 18 y = 10
Step-by-step explanation:
let the first number be x
let the second number be y
x = y + 8..... equation 1
2x + y = 46.... equation 2
x is the larger number
y is the smaller number.
Rearrange the equation and add equation 1 to equation 2.
x - y = 8
+ 2x + y = 46
-------------------
3x + 0 = 54
3x = 54
divide both sides by 3
x = 54/3
x = 18
Substitute x = 18 into equation 1
x = y + 8
18 = y + 8
collect like terms
y = 18-8
y = 10
Answer:
40 mph
Step-by-step explanation:
To find the maximum speed at which the car can travel, as the distance it requires to stop is 168 feet, we just need to use the value of d = 168 in the equation, and then find the value of s:
168 = 0.05s^2 + 2.2s
0.05s^2 + 2.2s - 168 = 0
Using Bhaskara's formula: we have:
Delta = 2.2^2 + 4*0.05*168 = 38.44
sqrt(Delta) = 6.2
s1 = (-2.2 + 6.2)/0.1 = 40 mph
s2 = (-2.2 - 6.2)/0.1 = -84 mph (a negative value does not make sense as 's' is the speed of the car)
So the maximum speed of the car is 40 mph
4.700 decimal form or maybe 4700 without decimal
