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77julia77 [94]
3 years ago
15

Mr. Bins sold 1/2 of his farm to Mr. Frost and 100 acres to Mr. Slate. He now has 1/5 of his original farm land. How big was his

farm originally? How much land does he have now?
Mathematics
1 answer:
kvasek [131]3 years ago
3 0

Answer:

Step-by-step explanation:

1/2

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A rectangular backyard measures 15 yards by 20 yards. There is a square flower bed with sides of 3 yards in the middle. What is
Delvig [45]
How to solve: (15x10)-(3x3)
Answer(hopefully is right): 291
6 0
3 years ago
Using distributive property what are two ways to find the product of 127 and 32?
timofeeve [1]
127 = 100 +27
32 = 30 +2
so 127 x 32 = (100 + 27) x (30+2) = (100x30) + (100x2) + (27x30) +(27x2)
= 3000+200+810+54 = 4064

second way:

127 = 100 + 27
127 x 32 = (100+27) x 32 = (100x 32) + ( 27x32) = 3200 + 864 = 4064
6 0
4 years ago
Find two numbers, if Their sum is 3/5 and their difference is 7 1/5
atroni [7]
Set up a system of equations:

x + y = 0.6
x - y = 7.2

You can solve using elimination. Add the two equations together:

2x + 0y = 7.8
2x = 7.8
x = 3.9

Then plug the number back into either of the two original equations to solve for y:

3.9 + y = 0.6
y = -3.3

The two numbers are 3.9 and -3.3
7 0
3 years ago
Read 2 more answers
On a particular map, 3 inches on the map equates to 10 miles in real life. If you know that the real-life distance between two b
Yakvenalex [24]

Answer:

15\dfrac{39}{40} $ inches

Step-by-step explanation:

On the map, 3 inches = 10 miles in real life.

Therefore:

10 miles: 3 Inches

1$ miles rep $ \dfrac{3}{10}$ inches

If the real-life distance between two buildings is 53.25 miles

Distance on the map

\dfrac{3}{10}$ X 53.25 inches\\=15.975 inches

=15\dfrac{39}{40} $ inches

The distance between the two buildings on the map is therefore: 15\dfrac{39}{40} $ inches

3 0
3 years ago
Determine the sum -46.38 (-24.6)
Zarrin [17]
Well, this equation as a sum would be -46.38 + -24.6 
so the answer would be -70.98. 

I hope this helps, and have a fantastic day!
8 0
3 years ago
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