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IgorC [24]
3 years ago
9

Find the slope for the line that passes thru the points (-2, 7) and (4, 18)

Mathematics
2 answers:
DedPeter [7]3 years ago
8 0

Answer:

m=\frac{11}{6}

General Formulas and Concepts:

<u>Pre-Alg</u>

  • Order of Operations: BPEMDAS

<u>Alg I</u>

  • Slope Formula: m=\frac{y_2-y_1}{x_2-x_1}

Step-by-step explanation:

<u>Step 1: Define</u>

Point (-2, 7)

Point (4, 18)

<u>Step 2: Solve for slope </u><em><u>m</u></em>

  1. Substitute:                    m=\frac{18-7}{4+2}
  2. Subtract/Add:               m=\frac{11}{6}
ra1l [238]3 years ago
5 0

Answer:

dkdkfdodffffdsddddddddd

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Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse
Drupady [299]

Answer:

a'(d) = \frac{d}{5} + \frac{3}{5}

a(a'(d)) = a'(a(d)) = d

Step-by-step explanation:

Given

a(d) = 5d - 3

Solving (a): Write as inverse function

a(d) = 5d - 3

Represent a(d) as y

y = 5d - 3

Swap positions of d and y

d = 5y - 3

Make y the subject

5y = d + 3

y = \frac{d}{5} + \frac{3}{5}

Replace y with a'(d)

a'(d) = \frac{d}{5} + \frac{3}{5}

Prove that a(d) and a'(d) are inverse functions

a'(d) = \frac{d}{5} + \frac{3}{5} and a(d) = 5d - 3

To do this, we prove that:

a(a'(d)) = a'(a(d)) = d

Solving for a(a'(d))

a(a'(d))  = a(\frac{d}{5} + \frac{3}{5})

Substitute \frac{d}{5} + \frac{3}{5} for d in  a(d) = 5d - 3

a(a'(d))  = 5(\frac{d}{5} + \frac{3}{5}) - 3

a(a'(d))  = \frac{5d}{5} + \frac{15}{5} - 3

a(a'(d))  = d + 3 - 3

a(a'(d))  = d

Solving for: a'(a(d))

a'(a(d)) = a'(5d - 3)

Substitute 5d - 3 for d in a'(d) = \frac{d}{5} + \frac{3}{5}

a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}

Add fractions

a'(a(d)) = \frac{5d - 3+3}{5}

a'(a(d)) = \frac{5d}{5}

a'(a(d)) = d

Hence:

a(a'(d)) = a'(a(d)) = d

7 0
3 years ago
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