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GarryVolchara [31]
4 years ago
5

A baker can bake 8 dozen cookies in 20 minutes. At this rate, how many minutes will it take to make 30 dozen cookies?

Mathematics
1 answer:
Viktor [21]4 years ago
4 0
It would take about an hour and 20 minutes to make 30 dozen cookies.
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I can’t figure this out please help me out
fomenos

Answer:

The answer is -1\4

Step-by-step explanation:

because its negative you can't expect it  to be positive 1\4 and its the only one that made sense

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3 years ago
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Jenny spends $1.15 a day for lunch. Her cafeteria lunch account has a balance of $36.80. Write an equation that can be used to d
ExtremeBDS [4]

Answer:

Balance Divided By Lunch Cost

Step-by-step explanation:

Divide $36.80 by $1.15 to receive the answer 32. She can eat for 32 days.

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3 years ago
I NEED HELP VOLUME OF SPHERES 5 QUESTIONS ASAP!!!!
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Answer:

1.)4188.79

2.) 7238.23

3.)1.02

4.)4189

5.)170 cm³

Step-by-step explanation:

1.) 4πr2 * 5

2.) 4πr2 * 12

3.) V = 4/3(PI*r3). *4.5

4.) V = 4/3 π r ^3 * 10

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3 balls each have a radius of 3 cm.

Volume of a cylinder = π r² h

V = 3.14 * (3cm)² * 18 cm

V = 508.68 cm³

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V = 113.04 cm³

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3 years ago
Eighteen boys joined a group of p students in the auditorium. If the ratio of boys to girls was then 5:4 write and algebraic exp
ycow [4]

Answer:  The number of girls in the auditorium is represented by the algebraic expression y=(4x +72) /5

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

  • <em>Number of boys added: 18 </em>
  • <em>Ratio of boys to girls: 5:4 </em>

So, the total number of boys is:

x + 18  

Number of girls = y

Number of boys / number of girls = 5/4

(x+18) /y = 5/4

Solving for "y"

4 (x +18) =5 y

4x +72 = 5y

(4x +72) /5 = y

The number of girls in the auditorium is represented by the algebraic expression y=(4x +72) /5

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3 years ago
Please help I really need it for a grade boost.
ArbitrLikvidat [17]

Answer:

I hope This will Help u.. Plz mark me as Brilliant please

Step-by-step explanation:

Before you get started, take this readiness quiz.

Simplify: 

If you missed this problem, review (Figure).

Solve Equations with Constants on Both Sides

In all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we will learn to solve equations in which the variable terms, or constant terms, or both are on both sides of the equation.

Our strategy will involve choosing one side of the equation to be the “variable side”, and the other side of the equation to be the “constant side.” Then, we will use the Subtraction and Addition Properties of Equality to get all the variable terms together on one side of the equation and the constant terms together on the other side.

By doing this, we will transform the equation that began with variables and constants on both sides into the form  We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

Solve: 

Solution

In this equation, the variable is found only on the left side. It makes sense to call the left side the “variable” side. Therefore, the right side will be the “constant” side. We will write the labels above the equation to help us remember what goes where.



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Use the Subtraction Property of Equality.Simplify.Now all the variables are on the left and the constant on the right.

The equation looks like those you learned to solve earlier.Use the Division Property of Equality.Simplify.Check:Let .

Solve: 



Solve: 



Solve: 

Solution

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3 years ago
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