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Basile [38]
3 years ago
10

Is the equation y+1=7(x+2) in point-slope form? justify your answer.​

Mathematics
1 answer:
djverab [1.8K]3 years ago
8 0
Yes it is because y-y1=slope(x-x1).
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Help please and thaank you
finlep [7]

Last one D

Step-by-step explanation:

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2 years ago
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What in the world???
vodomira [7]

Answer:

g²/f⁷h

Step-by-step explanation:

Because f⁹-f²=f⁷ which leaves f⁷ on the bottom, g³-g=g² which leaves g² on the top, and h⁵-h⁴=h which leaves h on the bottom.

8 0
3 years ago
A researcher decides to split scores on an exam into quartiles. She determines that a score of 64 is at the 25th percentile, a s
allochka39001 [22]

Answer:

Answer : A

Step-by-step explanation:

The given data is 25 th percentile is 64, 50th percentile is 74 and 75 th percentile is 80.

percentage : 25    50   75

  score        : 64    74    80

Median:- The median is obtained by first arranging the data in ascending or descending order and applying the following rule.

If the number of observations is odd, then the median is  observation

(\frac{n+1}{2}) ^{th} term

If the number of observations is even, then the median is  observation and  observations.

(\frac{n}{2}+1) ^{th}

given n=3, middle term is '74'

 In this given data the median is (M) = 74

Interquartile range IQR = median of upper half-median of lower half

                                       = 80-64

                                       = 16

                            IQR = 16

8 0
3 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
12=2(q-7)<br> Solve for q
RUDIKE [14]

Answer:

not sure just want points

Step-by-step explanation:

3 0
2 years ago
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