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tatuchka [14]
3 years ago
7

A. One day Annie weighed 24 ounces more than Benjie, and Benjie weighed 3 1/4 pounds less than Carmen. How did Annie’s and Carme

n’s weights compare on that day?
B. Why can’t you tell how much each person weighed?

Draw diagrams to support your answer.
Mathematics
1 answer:
Debora [2.8K]3 years ago
3 0
<h2>Answer:</h2>

A.

Let Annie's weight be = a

Let Benjie's weighs = b

Let Carmen's weight be = c

One day Annie weighed 24 ounces more than Benjie, equation forms:

a=b+24        ......(1)

Benjie weighed 3 1/4 pounds less than Carmen.  

In ounces:

1 pound = 16 ounces

\frac{13}{4} pounds = \frac{13}{4}\times16=52 ounces

b=c-52  or

c=b+52    ......(2)

Now adding (1) and (2), we get

a+b=b+24+c-52

=> a=c-28

This gives  Annie weighs 28 ounces less than Carmen.

B.

We cannot know anyone's actual weight, as we only know their relative weights.

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Determine whether the given linear equations are parallel, perpendicular, or neither.
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A. Perpendicular

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1/9  

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sveta [45]

This question was not written properly

Complete Question

Steve has 54 stamps in his collection of 13 cent, 29 cent, and 45 cent stamps, totaling a value of $15.98. If the number of 13 cent stamps is doubled, the new total value of his stamp collection would be $17.80. Find the number of each type of stamp in his collection.

Answer:

a) The number of 13 cent = 14

b) The number of 29 cent = 24

c) The number of 45 cent = 16

Step-by-step explanation:

We are told he has 54 stamps in total.

These stamps are;13 cent, 29 cent, and 45 cent stamps.

Let's

Number of 13 cent = a

Number of 29 cent = b

Number of 45 cent = c

Hence,

a + b + c = 54........... Equation 1

The total value of the 3 stamps are $15.98.

1 cent = $0.01

We have

0.13a + 0.29b + 0.45c= $15.98 ......... Equation 2

We know from the question that:

13 cent stamps is doubled, the new total value of his stamp collection would be $17.80.

Therefore

2(0.13a) + 0.29b + 0.45c = 17.8

0.26a + 0.29b + 0.45c = 17.8 ......... Equation 3

Combining Equation 2 and 3 together

0.13a + 0.29b + 0.45c= 15.98 ......... Equation 2

0.26a + 0.29b + 0.45c = 17.8 ......... Equation 3

We eliminate b and c by Subtracting Equation 3 from 3

0.13a = 1.82

a = 1.82/0.13

a = 14

a + b + c = 54........... Equation 1

Since a = 14

14 + b + c = 54

b + c = 54 - 14

b + c = 40

c = 40 - b

Substitute 40 - b for c and 14 for a is Equation 2

0.13a + 0.29b + 0.45c= 15.98 ......... Equation 2

0.13(14 ) + 0.29b + 0.45(40 - b) = 15.98

= 1.82 + 0.29b + 18 - 0.45b = 15.98

Collect like terms

1.82 + 18 - 15.98 = 0.45b - 0.29b

3.84 = 0.16b

b = 3.84/0.16

b = 24

Hence: Solving for c

a + b + c = 54........... Equation 1

14 + 24 + c = 54

38 + c = 54

c = 54 - 38

c = 16

Therefore,

The number of 13 cent = 14

The number of 29 cent = 24

The number of 45 cent = 16

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Answer:

7a + 5

Step-by-step explanation:

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