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nevsk [136]
3 years ago
13

I would appreciate it a lot

Mathematics
2 answers:
stich3 [128]3 years ago
5 0
The answer is C
I hope this helps you
Volgvan3 years ago
3 0

Answer:

C

Step-by-step explanation:

f(x) = Price

2.00 = Price of one ride

x = number of times ridden

f(x) = 2x

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Help answer 1 and 2 pls
Serjik [45]

Answer:

1. 0,4

2. x=-2

Step-by-step explanation:

5 0
3 years ago
How do you solve number 8 and what is its probability
Scorpion4ik [409]
Probably a 50% chance ??
5 0
3 years ago
How tall is the tower?
Karo-lina-s [1.5K]

Answer:

  • 93.6 ft

Step-by-step explanation:

\bf \cfrac{AB}{BC}  =  \tan(60)

  • \bf \cfrac{x}{50}  =  \sqrt{50}
  • \bf x = 50 \sqrt{3}

Length of tower:-

  • \bf x + 7
  • \bf 50 \sqrt{3}  + 7
  • \bf 93.6025
  • \bf = 93.6

Therefore, the tower is 93.6ft tall.

<u>-------------------------------------------</u>

5 0
2 years ago
The ratio of boys to girls in a class is 5:4. There are 36 Students in the class.How many students are girls
babymother [125]

Answer:

16 girls

Step-by-step explanation:

boys : girls : total

5          4        5+4 = 9

take the total number of people and divide by 9

36/9 = 4

Each number should be multiplied by 4

boys : girls : total

5*4        4*4     9*4

20          16        36

There are 16 girls

3 0
3 years ago
Read 2 more answers
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
3 years ago
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