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Vesna [10]
3 years ago
11

Dane's puppy eight 8 ounces when it was born. Now the puppy weighs 18 times as much as it did when it was born. How many pounds

does Dane's puppy weigh now. Please explain then give the answer.
Mathematics
2 answers:
Rzqust [24]3 years ago
7 0
First multiply 8 by 18 = 144. Then convert into pounds: 144÷16=9
Elanso [62]3 years ago
5 0
First you multiply the 8 by 18 because it tells you it weighs 18 more pounds then what it did before. You also need to multiply to get how many ounces he weighs in order to convert. When you multiply you get 144 ounces. Now your going to divide 144 by 16 because there are 16 ounce in 1 lb. You know ur dividing because ur converting from a smaller unit to a larger unit. When you divide you get 9 lbs.
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Joey has 5 collector stamps.He receives 3 more stamps each week that he completes community service hours.X=the number of weeks
kirza4 [7]

Answer:

Therefore the required equation is

Y=5+3X

Step-by-step explanation:

Given that , Joey has 5 collector stamps.

He gets 3 more stamps each week.

The number of week is X.

1st week he got 3 stamps

2nd week he got (3+3)=2.3 stamps

3rd week he got (3+3+3)=3.3 stamps

Stamp that he got=( the number of week ×3)

Therefore at end of Xth week he got =(X × 3) =3X stamps

Total number of stamps that collected after x week

=5+3X

Therefore the required equation is

Y=5+3X

6 0
4 years ago
Help asap give brain list
Ilia_Sergeevich [38]
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6 0
3 years ago
What is the common ratio for this geometric sequence?
juin [17]
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4 0
3 years ago
Read 2 more answers
In figure qr = 1/3pr. find pq​
BaLLatris [955]

Answer:

PQ = 3.2

Step-by-step explanation:

QR =\frac{1}{3} PR

QR =\frac{1}{3} \times 4.8

QR =1.6

PQ = PR - QR

PQ = 4.8 - 1.6

PQ = 3.2

7 0
3 years ago
Suppose the total area is represented by the equation x^2+14x+24. If the area 1 is x^2 square units and area IV is 24 units, wha
Nutka1998 [239]

Answer:

See Explanation

Step-by-step explanation:

Given

Total = x^2 + 14x + 24

Area\ I = x^2

Area\ IV = 24

Required

True about area II and III

The question is incomplete; so, I will answer using general knowledge.

If the total area is:

Total = x^2 + 14x + 24

And there are 4 areas, then;

Total = Area\ I + Area\ II + Area\ III + Area\ IV

Rewrite as:

Total = Area\ I + Area\ IV+ Area\ II + Area\ III

This gives:

x^2 + 14x + 24 = x^2 + 24+ Area\ II + Area\ III

Collect like terms

x^2 -x^2 + 14x + 24 -24=  Area\ II + Area\ III

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Rewrite as:

Area\ II + Area\ III  = 14x

<em>So, the sum of areas III and IV must be 14x</em>

4 0
3 years ago
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