Aaron is making 3 1/2 batches of brownies. Each batch takes 2 1/2 cups of flour. How much flour will John need for the brownies?
1 answer:
You might be interested in
Answer:
Option B
Step-by-step explanation:
A unit circle means radius of the circle = 1 unit
Let a terminal point on the circle is (x, y) and angle between the point P and x-axis is θ.
Center of the circle is origin (0, 0).
Therefore, ordered pair representing the terminal point will be (OP×Cosθ, OP×Sinθ) =
OP.Cosθ = 1×Cosθ =
Cosθ =
θ = ,
where n = integers
Similarly, OP×Sinθ = 1×Sinθ = -
Sinθ = -
θ = ,
where n = integer
Common value of θ will be, θ =
Option B will be the answer.
Answer:
y=-0.215x^2+35
Step by Step:
Let, ,
,
,
We know that, the general equation of the parabola.
Substitute the value of in equation
and find the value of
Hence, the equation of the parabola is:
3.) An extreme value refers to a point on the graph that is possibly a maximum or minimum. At these points, the instantaneous rate of change (slope) of the graph is 0 because the line tangent to the point is horizontal. We can find the rate of change by taking the derivative of the function.
y' = 2ax + b
Now that we where the derivative, we can set it equal to 0.
2ax + b = 0
We also know that at the extreme value, x = -1/2. We can plug that in as well.
The 2 and one-half cancel each other out.
Now we know that a and b are the same number, and that ax^2 + bx + 10 = 0 at x = -1/2. So let's plug -1/2 in for x in the original function, and solve for a/b.
a(-0.5)^2 + a(-0.5) + 10 = 0
0.25a - 0.5a + 10 = 0
-0.25a = -10
a = 40
b = 40
To determine if the extrema is a minima or maxima, we need to go back to the derivative and plug in a/b.
80x + 40
Our critical number is x = -1/2. We need to plug a number that is less than -1/2 and a number that is greater than -1/2 into the derivative.
LESS THAN:
80(-1) + 40 = -40
GREATER THAN:
80(0) + 40 = 40
The rate of change of the graph changes from negative to positive at x = -1/2, therefore the extreme value is a minimum.
4.) If the quadratic function is symmetrical about x = 3, that means that the minimum or maximum must be at x = 3.
y' = 2ax + 1
2a(3) + 1 = 0
6a = -1
a = -1/6
So now plug the a value and x=3 into the original function to find the extreme value.
(-1/6)(3)^2 + 3 + 3 = 4.5
The extreme value is 4.5
y' = 2ax + b
Now that we where the derivative, we can set it equal to 0.
2ax + b = 0
We also know that at the extreme value, x = -1/2. We can plug that in as well.
The 2 and one-half cancel each other out.
Now we know that a and b are the same number, and that ax^2 + bx + 10 = 0 at x = -1/2. So let's plug -1/2 in for x in the original function, and solve for a/b.
a(-0.5)^2 + a(-0.5) + 10 = 0
0.25a - 0.5a + 10 = 0
-0.25a = -10
a = 40
b = 40
To determine if the extrema is a minima or maxima, we need to go back to the derivative and plug in a/b.
80x + 40
Our critical number is x = -1/2. We need to plug a number that is less than -1/2 and a number that is greater than -1/2 into the derivative.
LESS THAN:
80(-1) + 40 = -40
GREATER THAN:
80(0) + 40 = 40
The rate of change of the graph changes from negative to positive at x = -1/2, therefore the extreme value is a minimum.
4.) If the quadratic function is symmetrical about x = 3, that means that the minimum or maximum must be at x = 3.
y' = 2ax + 1
2a(3) + 1 = 0
6a = -1
a = -1/6
So now plug the a value and x=3 into the original function to find the extreme value.
(-1/6)(3)^2 + 3 + 3 = 4.5
The extreme value is 4.5