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Ilia_Sergeevich [38]

2 years ago

7

The foodservice department has been asked to prepare a special luncheon for the grandway hospital board of directors on tuesday.

there will be 25 board members at the luncheon. there are 125 portions of vegetable lasagna needed for the hospital's lunch service that day. the standardized vegetable lasagna recipe yields 125 portions and requires 21 lb of ricotta cheese. how much ricotta cheese must chef carl order to meet production needs?

1 answer:

AURORKA [14]2 years ago

5 0

125/21 = 150/x
cross multiply
125x = 3150
x = 3150/125
x = 25.2 lbs....rounded up = 26 lbs <==

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Fofino [41]

Answer:

The dimensions of the two different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

The perimeter of the two different rectangular regions are;

1st Arrangement:

P₁ = 18 yards

2nd Arrangement:

P₂ = 24 yards

Step-by-step explanation:

Bella is putting down patches of sod to start a new lawn.

She has 20 square yards of sod.

We are asked to provide the dimensions of two different rectangular regions that she can cover with the sod.

Recall that a rectangle has an area given by

Area = W*L

Where W is the width of the rectangle and and L is the length of the rectangle.

Since Bella has 20 square yards of sod,

20 = W*L

There are more than two such possible rectangular arrangements.

Out of them, two different possible arrangements are;

1st Arrangement:

20 = (4)*(5) = (5)*(4)

Width is 4 yards and length is 5 yards or width is 5 yards and length is 4 yards

2nd Arrangement:

20 = (2)*(10) = (10)*(2)

Width is 2 yards and length is 10 yards or width is 10 yards and length is 2 yards

Therefore, the dimensions of two different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

What is the perimeter of each region?

The perimeter of a rectangular shape is given by

P = 2(W + L)

Where W is the width of the rectangle and and L is the length of the rectangle.

The perimeter of the 1st arrangement is

P₁ = 2(4 + 5)

P₁ = 2(9)

P₁ = 18 yards

The perimeter of the 2nd arrangement is

P₂ = 2(2 + 10)

P₂ = 2(12)

P₂ = 24 yards

So the perimeter of the 1st arrangement is 18 yards and the perimeter of the 2nd arrangement is 24 yards.

Note:

Another possible arrangement is,

20 = (1)*(20) = (20)*(1)

Width is 1 yard and length is 20 yards or width is 20 yards and length is 1 yard.

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1+7 and 7+1 are the same equations. The numbers are just switched around .
Example:
1+2=3
2+1+3

<span>They add up to the same answer no matter where they are placed, therefore knowing 1+7 helps you find the sum of 7+1 (again, because they are the same) </span>

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Answer

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Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

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Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

<u>Josh:</u>

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em> $h(t)=0$ <em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

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Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

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Answer:not A

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