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2 years ago

9. Write the equation of the line in slope-intercept form that has the following points: (2, -1)(5,-3)

Mathematics

1 answer:

Crazy boy [7]2 years ago

3 0

Answer:

$y = - \frac{2}{ 3} x + \frac{1}{3}$

Step-by-step explanation:

either pick the generic line in slope-intercept form

$y = mx + q$

and plug y and x from the points in there, you get two conditions on m and q and find the values or...

Between the two points you get that when x increases (from 2 to 5) by 3 units, y decreases (from -1 to -3) by 2 units. You can rule out the first two options since the slope cannot be an integer. from the other 2, try x=5 into both to see which one gives -3 for y.

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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

Need help on writing a proof.

viva [34]

Answer: Congruent

Step-by-step explanation:

Line BC is reflexive to itself

AB is congruent to CD, and that's given

Line AC is congruent to BC, using these prior proofs

Angle EDC and Angle FAB are congruent, and that's a given.

You can conclude that they are congruent by AAS

5 0

3 years ago

Bruno uses a piece of wrapping paper with dimensions 1 and one-fourth feet by 3 feet to wrap a gift. What is the total area of t

8_murik_8 [283]

The area of the paper used to wrap the gift is 3 3/4 square feet

<h3>How to determine the total area?</h3>

The dimension is given as:

1 1/4 by 3

The area is the product of the dimensions.

So, we have

Area = 1 1/4 * 3

Express as an improper fraction

Area = 5/4 * 3

Evaluate

Area = 15/4

Rewrite as:

Area = 3 3/4

Hence, the area of the paper used to wrap the gift is 3 3/4 square feet