Student B:
1/2 / 4
= 1/2 x 1/4
= 1/8
student A:
2/3 / 6
= 2/3 x 1/6
= 1/9
student C:
4/5/ 8
= 4/5 x 1/8
= 1/10
answer
student B: 1/8 of a mile per minute
student A: 1/9 of a mile per minute
student C: 1/10 of a mile per minute
Answer:
B:0.2(48)
Step-by-step explanation:
Of means multiply
20% of 48
Changing to decimal form
.20 * 48
The area of the triangle is
A = (xy)/2
Also,
sqrt(x^2 + y^2) = 19
We solve this for y.
x^2 + y^2 = 361
y^2 = 361 - x^2
y = sqrt(361 - x^2)
Now we substitute this expression for y in the area equation.
A = (1/2)(x)(sqrt(361 - x^2))
A = (1/2)(x)(361 - x^2)^(1/2)
We take the derivative of A with respect to x.
dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]
dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]
Now we set the derivative equal to zero.
(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0
-2x^2 + 361 = 0
-2x^2 = -361
2x^2 = 361
x^2 = 361/2
x = 19/sqrt(2)
x^2 + y^2 = 361
(19/sqrt(2))^2 + y^2 = 361
361/2 + y^2 = 361
y^2 = 361/2
y = 19/sqrt(2)
We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
Answer:
80
Step-by-step explanation:
If you add each value it equals 30.
30x10=300 so 8 (value of the 9th grade bar) times 10 equals 80.
Answer:
Ummmm
No?
Step-by-step explanation: