Answer:
![h](https://tex.z-dn.net/?f=h%3C8)
Step-by-step explanation:
If Bradley makes less than $8 per hour.
Let Bradley's hourly rate = h
Therefore, an inequality to represent the real-world problem is:
![h](https://tex.z-dn.net/?f=h%3C8)
- The variable in the inequality represents <u>Bradley's hourly rate</u>
- The inequality symbol needed is <u>Less Than</u>
- The constant is <u>8</u>
- The inequality that correctly represents the problem is:
![h](https://tex.z-dn.net/?f=h%3C8)
Hello from MrBillDoesMath!
Answer:
27 x^2 sqrt(x^3) sin(x^3) - 9/2 sqrt(x^1/2) sin(x^1/2)*x^(-1/2)
Discussion:
Let f(t) - 9 sqrt(t) sin(t), then
y' = f(x^3) * d(x^3)/dx - f(sqrt(x)) * d(x^1/2)/dx
= (9 sqrt(x^3)sin(x^3)) * 3x^2 - (9 sqrt(x^1/2)sin(x^1/2)) * (1/2) x^-(1/2)
= 27 x^2 sqrt(x^3) sin(x^3) - 9/2 sqrt(x^1/2) sin(x^1/2)*x^(-1/2)
Hope I didn't make a "bozo" error differentiating things!
Thank you,
MrB
Answer:
The value of y = 30 when x = 5
Step-by-step explanation:
We know that when 'y' varies directly proportional to 'x', we get the equation
y ∝ x
y = kx
k = y/x
where k is called the 'constant of direct variation'.
We are given
y = 12 when x = 2
so substituting y = 12 and x = 2 in the equation
k = y/x
k = 12/2
k = 6
Thus, the value of k = 6
We have to determine the value of y when x = 5
so substitute x = 5 and k = 6 in the equation
y = kx
y = 6(5)
y = 30
Therefore, the value of y = 30 when x = 5
Answer:
y=-2/5x+5
Step-by-step explanation:
Answer:
-2y(y + 4)
Step-by-step explanation:
-10y^2 - 3y - (5y - 8y^2)
= -10y^2 - 3y - 5y + 8y^2
= -2y^2 - 8y
= -2y(y + 4)