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1)

a raccoon, a cat, and a fox have weights in the ratios 5:3:8 respectively. If their combined weights is 48 lb how heavy is the raccoon?

2) and 3)

Guest Apr 17, 2017

#1
+7340
+3

1.

Let's call the unknown scale factor "s".

Now we can make this equation:

5s + 3s + 8s = 48 lb

16s = 48 lb

s = 48/16 lb

s = 3 lb

And the weight of the raccoon = 5s = 5*3 = 15 lb

2.

Let's call the number of nickles "n" and the number of dimes "d".

Now we can make these equations:

n + d = 48

5n + 10d = 390

Solve the first one for n (or d) and substitute.

n = 48 - d

5(48 - d) + 10d = 390

240 - 5d + 10d = 390

5d = 390 - 240

d = 150/5 = 30

n = 48 - 30 = 18

3.

Let's call the number of Mommy fish "y" and the number of Mama fish "a".

Now we can make these equations:

y + a = 10

220y + 480a = 2460

Solve the first equation for y (or a) and substitute.

y = 10 - a

220(10 - a) + 480a = 2460

2200 - 220a + 480a = 2460

260a = 2460 - 2200

a = 260/260 = 1

y = 10 - 1 = 9

There were 9 Mommy fish, so there were 220*9 = 1980 Mommy fish eggs, just like you found! :)

hectictar  Apr 18, 2017
#1
+7340
+3

1.

Let's call the unknown scale factor "s".

Now we can make this equation:

5s + 3s + 8s = 48 lb

16s = 48 lb

s = 48/16 lb

s = 3 lb

And the weight of the raccoon = 5s = 5*3 = 15 lb

2.

Let's call the number of nickles "n" and the number of dimes "d".

Now we can make these equations:

n + d = 48

5n + 10d = 390

Solve the first one for n (or d) and substitute.

n = 48 - d

5(48 - d) + 10d = 390

240 - 5d + 10d = 390

5d = 390 - 240

d = 150/5 = 30

n = 48 - 30 = 18

3.

Let's call the number of Mommy fish "y" and the number of Mama fish "a".

Now we can make these equations:

y + a = 10

220y + 480a = 2460

Solve the first equation for y (or a) and substitute.

y = 10 - a

220(10 - a) + 480a = 2460

2200 - 220a + 480a = 2460

260a = 2460 - 2200

a = 260/260 = 1

y = 10 - 1 = 9

There were 9 Mommy fish, so there were 220*9 = 1980 Mommy fish eggs, just like you found! :)

hectictar  Apr 18, 2017