Answer:
Step-by-step explanation:
The volume of a cube can be found with this formula:
Where "s" is the lenght of any edge of the cube.
The formula for calculate the volume of a rectangular prism is:
Where "l" is the lenght, "w" is the width and "h" is the height.
We need to find the volume of a cube box:
To find the volume of the shipping box, first we must convert the mixed number to an improper fraction:
Then the volume of the shipping box is:
Now, in order to find the number of cube boxes can Haley fits into a shipping box, you must divide the the volume of the shipping box by the volume of one cube. This is:
Answer:
see explanation
Step-by-step explanation:
A fraction is in simplest form when no other factor but 1 will divide into the numerator/ denominator
(1)
← is in simplest form
(2)
← is in simplest form
(3)
( divide numerator/ denominator by 2 )
= ← in simplest form
Answer:
(6)
<u>Isosceles trapezoid</u>
- ∠R and ∠S are supplementary
- ∠S≅∠T, ∠R≅∠V
- m∠S = m∠T = 54°
- m∠R = 180° - 54° = 126°
(9)
<u>Midsegment is half the sum of the base lengths</u>
- ST = 1/2(FP + EQ)
- 3.1 = 1/2(2.7 + EQ)
- 2.7 + EQ = 6.2
- EQ = 6.2 - 2.7
- EQ = 3.5
Let x represent the number of pounds of $10.75/lb coffee used in the mix.
.. 10.75x +3.25(100 -x) = 5.65*100 . . . . . . total cost of coffee blend
.. x(10.75 -3.25) = 100*(5.65 -3.25) . . . . . .separate x-terms from non-x terms
.. x = 240/7.5 = 32
32 pounds of $10.75/lb coffee should be used
68 pounds of $3.25/lb coffee should be used
now, we get critical points from zeroing out the derivative, and also from zeroing out the denominator, but those at the denominator are critical points where the function is not differentiable, namely a sharp spike or cusp or an asymptote.
so, from zeroing out the derivative we get no critical points there, from the denominator we get x = 8, but can't use it because f(x) is undefined.
therefore, we settle for the endpoints, 4 and 6,
f(4) =3 and f(6) = 7
doing a first-derivative test, we see the slope just goes up at both points and in between, but the highest is f(6), so the absolute maximum is there, while we can take say f(4) as the only minimum and therefore the absolute minumum as well.