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amid [387]
3 years ago
11

Find the center and radius of x^2+y^2+2x-10y+10=0

Mathematics
1 answer:
kumpel [21]3 years ago
3 0
r^{2} = 16
r=4
(xh)^{2}2+(y-k)2=(x+1)2 +(y-5)2
h=-1,k=5
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Two friends sold many pieces of furniture and made ​$1550 during their garage sale. They had fifteen more​ $10 bills than​ $50 b
Novosadov [1.4K]

Answer:

The number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.

Step-by-step explanation:

Let n₁ = number of $ 10 bills,

Let n₂ = number of $ 20 bills and

Let n₃ = number of $ 50 bills

Given that 10n₁ + 20n₂ + 50n₃ = 1550    (1)

Also, since we have fifteen more​ $10 bills than​ $50 bills, n₁ = n₃ + 15  (2)

and we have 4 more than three times as many​ $20 bills as​ $50 bills. n₃ = 3n₂ + 4. (3)

substituting equations (2) and (3) into (1), we have

10(n₃ + 15) + 20n₂ + 50n₃ = 1550

expanding the bracket, we have

10n₃ + 150 + 20n₂ + 50n₃ = 1550

collecting like terms, we have

60n₃ + 150 + 20n₂ = 1550

inserting equation (3), we have

60(3n₂ + 4) + 150 + 20n₂ = 1550

expanding the bracket, we have

180n₂ + 240 + 150 + 20n₂ = 1550

collecting like terms, we have

200n₂ + 390 = 1550

subtracting 390 from both sides, we have

200n₂ = 1550 - 390

200n₂ = 1160

dividing both sides by 200, we have

n₂ = 1160/200

n₂ = 5.8

n₂ ≅ 6 since it cannot be a fraction.

Substituting this into (3), we have

n₃ = 3n₂ + 4 = 3(6) + 4 = 18 + 4 = 22

substituting n₃ into (2), we have

n₁ = n₃ + 15 = 22 + 15 = 37

So, the number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.

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I think 8 is again maybe its true
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Answer:

6>x<10

Step-by-step explanation:

also put a line under both > <

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Solve. 45 + (-5) - 4 + 7<br> A: 45<br> B: 0<br> C: -34<br> D: 43
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When you add negative, it’s the same as subtracting
45 - 5 - 4 + 7
40 - 4 + 7
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The answer is D. 43
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60,000, 900,000, 302,000, 11,000

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