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mariarad [96]
3 years ago
5

At Yankee Candle, four large candles and three small candles sell for $34. For

Mathematics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

the answer is D

Step-by-step explanation:

{4x+3y=34

{2x+y=16

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Step-by-step explanation:

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3 years ago
On tuesday, donovan earned $11 for $2 of babysitting. on saturday, he babysat for the same family and earned $38.50. how many ho
Sedbober [7]
If its going by $11 for 2 hours then he babysat ?


ok so $11= 2 hours


11 x 3= 33    half of eleven is 5.5


so 33+ 5.5 = $38.50 so I believe the answer is 6 hours and 30 min but I'm not 100 percent positive
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3 years ago
Ecuacion de <br>3.4x+2=22.4​
lidiya [134]
Answer/respuesta:x=6

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jolli1 [7]

Answer:

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Step-by-step explanation:

4 0
2 years ago
What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
2 years ago
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