Answer:
<h2>There remains 7/6 of pie.</h2>Step-by-step explanation:
Givens
The remaining parts are 1/3 of one pie and 5/6 of the other pie.
To find how much pie is remaining, we just need to sum those fractions.
Therefore, there remains 7/6 of pie. They ate just 1/6.
Answer:
The statements 1 and 4 are true.
Step-by-step explanation:
In order to quickly visualize the center and radius of the circle we can manipulate the given equation in such a way that it changes to the standard form.
Now that the equation is its standard form we can compare it to the standard form formula to determine the center and radius of the circle:
Where are the coordinates of the center of the circle and
is the radius of the circle. Therefore the circle we were given has center in (1,0) and radius
. With this in mind we can analyze each statement.
1. The radius of the circle is 3 units. This statement is correct.
2. The center of the circle lies on the y-axis. This statement is false, since the center of the circle has the coordinates (1,0) it means that it actually lies in the x-axis.
3. The standard form of the equation is (x - 1)² + y² = 3. This is false the correct standard equation is (x - 1)² + y² = 9.
4. The radius of the circle is the same as the radius of the circle who's equation is x² + y² = 9. This statement is correct, the radius is given by the square root of the constant on the right side of the equation, therefore the radius of both circle is 3.
Answer:
Step-by-step explanation:
Represent the volume of the box with V and the dimensions with l, b and h.
The volume (V) is:
Make h the subject of the formula
The surface area (S) of the aquarium is:
Where lb represents the area of the base (i.e. slate):
The cost (C) of the surface area is:
Substitute for h in the above equation
Differentiate with respect to l and with respect to b
To solve for b and l, we equate both equations and set l to b (to minimize the cost)
By comparison:
becomes
Cross Multiply
Solve for l
Recall that:
Also recall that:
Apply law of indices
The dimension that minimizes the cost of material of the aquarium is: