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Marrrta [24]
3 years ago
9

3(x+5x-5) another give away rate me...

Mathematics
1 answer:
makkiz [27]3 years ago
7 0

Answer:

18x-15

Step-by-step explanation:

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(6, -12), (15,-3) <br> slopeee
zzz [600]
The answer is
M=1
Your welcome:)
6 0
3 years ago
Three friends earned more than $200 washing cars. They paid their parents $28 for supplies and divided the rest of the money equ
Ksenya-84 [330]

Answer: 3x + 28 > 200

Step-by-step explanation:

$$Let $x$ be the amount each friend received. Since there are 3 friends, then $3 x$ is the amount of money they split evenly. The amount of money they split evenly was the amount left over after paying their parent's $\$ 28$. Therefore $3 x+28$ is the total amount they earned.

7 0
2 years ago
Find the missing value in the proportion 12/34 = 18/m.
qaws [65]

Both the top and bottom numbers would have the same scale (ratio) in order to be a proportion.

Find the scale of the top numbers:

18/12 = 1.5

The top numbers have a scale of 1.5 so the bottom numbers also have a scale of 1.5

Multiply the known bottom number by the scale:

M = 34 x 1.5 = 51

7 0
3 years ago
Read 2 more answers
Who is the best team in the NFL
Stells [14]
<h3><u><em>Answer:</em></u></h3><h3><em>New England Patriots 6-4</em></h3>

6 0
2 years ago
Read 2 more answers
2^{51} mod 22 in words, two to the power of fifty-one mod twenty-two
andreyandreev [35.5K]

Since 2⁵ = 32, and

2⁵ ≡ 32 ≡ 10 (mod 22),

we have

2⁵¹ ≡ 2 • 2⁵⁰ ≡ 2 • (2⁵)¹⁰ ≡ 2 • 10¹⁰ (mod 22)

Now consider 10¹⁰ (mod 22):

10 = 2 • 5

10¹⁰ ≡ 2¹⁰ • 5¹⁰ ≡ (2⁵)² • 5¹⁰ ≡ 10² • 5¹⁰ ≡ 2² • 5¹² (mod 22)

so that

2⁵¹ ≡ 2³ • 5¹² (mod 22)

Now consider 5¹² (mod 22):

5 and 22 are coprime, and ɸ(22) = 10 (where ɸ(<em>n</em>) is the Euler totient function). By Euler's theorem,

5¹² ≡ 5² • 5¹⁰ ≡ 5² • 1 ≡ 25 ≡ 3 (mod 22)

and so

2⁵¹ ≡ 2³ • 3 ≡ 24 ≡ 2 (mod 22)

Another, more tedious method: Start with smaller powers of 2 and look for a pattern.

2 ≡ 2 (mod 22)

2² ≡ 4 (mod 22)

2³ ≡ 8 (mod 22)

2⁴ ≡ 16 (mod 22)

2⁵ ≡ 32 ≡ 10 (mod 22)

2⁶ ≡ 2 • 32 ≡ 2 • 10 ≡ 20 (mod 22)

2⁷ ≡ 2 • 20 ≡ 40 ≡ 18 (mod 22)

2⁸ ≡ 2 • 18 ≡ 36 ≡ 14 (mod 22)

2⁹ ≡ 2 • 14 ≡ 28 ≡ 6 (mod 22)

2¹⁰ ≡ 2 • 6 ≡ 12 (mod 22)

2¹¹ ≡ 2 • 12 ≡ 24 ≡ 2 (mod 22)

2¹² ≡ 2 • 2 ≡ 4 (mod 22)

and so on, with a cyclic pattern of length 10. That is, 2^{10k+1}\equiv2\pmod{22} for any integer <em>k</em> ≥ 0. So 2⁵¹ ≡ 2 (mod 22).

3 0
3 years ago
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