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DENIUS [597]
4 years ago
10

sam had 70 erasers and rulers. after giving away 1/3 of his erasers and 10 rulers, he had an equal number of erasers and rulers

left. how many erasers did he have in the beginning?
Mathematics
1 answer:
Radda [10]4 years ago
4 0

Answer:

36 erasers

Step-by-step explanation:

Let number of erasers be e

let number of rulers be r

We can write:

e + r = 70

and

After giving away, he has

Erasers:  2/3e

Rulers:  r - 10

These two are equal, so we can write and solve:

2/3e = r - 10

2/3e + 10 = r

Putting this in initial equation, we have:

e + (2/3e + 10) = 70

5/3e + 10 = 70

5/3 e = 60

e = 36

And rulers is:

r = 2/3(36) + 10 = 34

Hence, he had 36 erasers in the beginning

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Answer:

Part 1) y+8=3(x+2)^{2} -----> y=-8.08

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we know that

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y-6=-(x-1)^{2}

the vertex is the point (1,6)      

\frac{1}{4p}=-1

p=-\frac{1}{4}    

The directrix is equal to

y=6+\frac{1}{4}=6.25

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