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kkurt [141]
3 years ago
11

To make 1 1/2 dozen muffins, a recipe uses 3 1/2 cups of flour. How many cups of flour are needed for every dozen muffins made?

Enter your answer in the box as a mixed number in simplest form.
Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
6 0

Answer: There are 2\dfrac{1}{3} cups of flour needed.

Step-by-step explanation:

Since we have given that

Number of dozen muffins = 1\dfrac{1}{2}=\dfrac{3}{2}

Number of cups of flour = 3\dfrac{1}{2}=\dfrac{7}{2}

So, we need to find the number of cups of flour are need for every dozen muffins made.

Number of cups of flour is given by

\dfrac{7}{2}\div \dfrac{3}{2}\\\\=\dfrac{7}{2}\times \dfrac{2}{3}\\\\=\dfrac{7}{3}\\\\=2\dfrac{1}{3}

Hence, there are 2\dfrac{1}{3} cups of flour needed.

ch4aika [34]3 years ago
5 0
Ratio and proportion

convert to inproper fracton first

1 and 1/2=1+1/2=2/2+1/2=3/2

3 and 1/2=3+1/2=6/2+1/2=7/2


3/2dozen=7/2cups
times 2 both sides
3dozen=7cups
divide by 3 both sides
1dozen=7/3cups=2 and 1/3 cup
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The sum of three consecutive natural numbers is 1251 find the numbers
valentinak56 [21]

Numbers are 416, 417 and 418

<u>Step-by-step explanation:</u>

Step 1:

Let the numbers be x, x + 1 and x + 2. Given that their sum is 1251.

⇒ x  + x + 1 + x + 2 = 1251

⇒ 3x + 3 = 1251

⇒ 3x = 1248

⇒ x = 416

Step 2:

Find the other numbers.

⇒ x + 1 = 417 and x + 2 = 418

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PLEASE HELP ME 16 POINTS PLEASE HELP
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Answer:

A.

Step-by-step explanation:

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Erica collected a total of $180 for her favorite charity over the first 3 months of the year. She collected the same amount in t
ankoles [38]
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HELP ASAP PLZZZZ
Tcecarenko [31]
QUESTION 1

The given system of equations is

3d - e = 7...eqn(1)
d + e = 5...eqn(2)

To solve by linear combination, we add equation (1) to equation (2) to get,

3d  + d= 7 + 5


4d = 12


We divide through by 4 to obtain,


d =  \frac{12}{4}


d = 3


We put d=3 into equation (2) to get,



3+ e = 5


e = 5 - 3


e = 2


\boxed {The \: solution \: is  \: (3, 2)}



QUESTION 2


The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)


To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

4x - 3x = 5 - 3



This implies that,

x = 2


Put x=3 into equation (1) to get,

4(2) + y = 5

8+ y = 5


y = 5 - 8



y =  - 3

The solution is

(2,-3)



QUESTION 3

We want to solve the system;


a – 2b = –2 ....eqn(1)


2a + 2b = 14...eqn(2)

by linear combination.


We need to add equation (1) to equation (2) to eliminate b.


This implies that,

2a + a = 14 +  - 2




Simplify,

3a = 12



Divide both sides by 3 to get,


a = 4
Put a=4 into equation (2) to obtain,



2(4) + 2b = 14


8 + 2b = 14
2b = 14 - 8


2b = 6


b = 3


The ordered pair in the form (a, b) is

(4,3)



QUESTION 4

The given system of equations is


11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)


We subtract equation (2) from equation (1) to get,


11x - 3x = 18 - 2


8x = 16


x = 2


Put x=2 into equation (2) to obtain,


3(2) + 4y = 2


This implies that,


6 + 4y = 2


4y = 2 - 6


4y =  - 4


y=-1

The correct answer is (2,-1).




QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2


We add the two equations to eliminate e.


This implies that,

2d + d = 8 + 4


3d = 12



We divide both sides by 3 to get,


d = 4


We put d=4 into equation (2) to get,

4 - e = 4

- e = 4 - 4



- e = 0



e = 0


The solution is

(4,0)
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3 years ago
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