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3 years ago

13

How many servings can be made from 108 cups of punch?

1 answer:

yarga [219]3 years ago

5 0

38 servings I think

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If 1900 square centimeters of material are available to make a box with a square base and an open top, find the largest possible

Evgen [1.6K]

Answer:

Volume = $7969$ cubic centimeter

Step-by-step explanation:

Let the length of each side of the base of the box be A and the height of the box be H.

Area of material required to make the box is equal to is $A^2 + 4*A*H.$

$A^2 + 4*A*H = 1900$

Rearranging the above equation, we get -

$`H = \frac{(1900 - A^2)}{(4*A)}$

Volume of box is equal to product of base area of box and the height of the box -

$V = A*A* H$

Substituting the given area we get -

$\frac{A^2*(1900 - A^2)}{4A} = \frac{(1900*A - A^3)}{4}$

For maximum volume

$\frac{dV}{dA} =0$

$\frac{ 1900}{4} - \frac{3*A^2}{4} = 0$

$A^2 = \frac{1900}{3}$

Volume of the box

= $\frac{\frac{1900}{3}*(1900 - \frac{1900}{3}) }{4 * \sqrt{\frac{1900}{3} } }$

= $7969$ cubic centimeter

3 0

2 years ago

On Marshal’s 5th-grade math test, he was asked to evaluate: 25,328.941 ÷ 104. The answer he wrote was 25.328941. Unfortunately f

dimulka [17.4K]

Answer:

25,328.941 ÷ 104= 7.90415302

Step-by-step explanation:

3 1

3 years ago

Your favorite basketpall player made two-point duele goals and three-point field goals for a total of 34 points. He made 16 shot

Ira Lisetskai [31]

Answer:

it was just a joke chilll

3 0

3 years ago

Please help me how many times can 3 go into 77

faust18 [17]

Answer:

3 can go into 77 25 times.

3 0

3 years ago

Find an upper limit for the zeroes 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0

erik [133]

<u>Answer-</u>

2 is the upper limit for the zeros.

<u>Solution-</u>

The given function f(x) is,

$2x^4 -7x^3 + 4x^2 + 7x - 6 = 0$

For calculating the zeros,

$\Rightarrow f(x)=0$

$\Rightarrow 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0$

$\Rightarrow 2x^4-4x^3-3x^3+6x^2-2x^2+ 4x+3x-6=0$

$\Rightarrow 2x^3(x-2)-3x^2(x-2)-2x(x-2)+3(x-2)=0$

$\Rightarrow (x-2)(2x^3-3x^2-2x+3)=0$

$\Rightarrow (x-2)(x^2(2x-3)-1(2x-3))=0$

$\Rightarrow (x-2)(x^2-1)(2x-3)=0$

$\Rightarrow (x-2)(x+1)(x-1)(2x-3)=0$

$\Rightarrow x-2=0,\ x+1=0,\ x-1=0,\ 2x-3=0$

$\Rightarrow x=2,\ x=-1,\ x=1,\ x=\frac{3}{2}$

From all the 4 roots, it can be obtained that 2 is the greatest zero.

7 0

2 years ago