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Temka [501]
3 years ago
15

Q6.) Write the​ slope-intercept equation of the function f whose graph satisifies the given conditions.

Mathematics
2 answers:
Lady_Fox [76]3 years ago
5 0
3x-7y-14 = 0

7y = 3x-14

y = (3/7)x - 2

slope of perpendicular line = -1/(3/7) = -7/3

so function f which is perpendicular and has same y-intercept is

y = (-7/3)x - 2

Brums [2.3K]3 years ago
3 0

The general form for a straight line is y = mx + c where m is slope and c is y-intercept.

Two lines are perpendicular when the product of their slopes = -1

Given line 3x - 7y - 14 = 0, its slope = 3/7 and y-intercept = -14/7 = -2

So a line perpendicular to the given and has the same y-intercept is

y = (-7/3) x - 2


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dmitriy555 [2]

Answer:

x=3

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Step-by-step explanation:

Substitution Method:

Substitute the value of x into the equation

2(6y-5)-3y=2

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7 0
4 years ago
Answer for both boxes please :)​
Jet001 [13]

Answer:

Concept: Linear systems

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7 0
3 years ago
Mariah swam her leg of the race in 24.98 seconds and Janet swam her leg of the relay race in 25.874 seconds. What was their comb
Svetach [21]

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4 0
4 years ago
Read 2 more answers
The product of an irrational number and a rational number is irrational. Sometimes True Always True Never True
shtirl [24]

Answer:

  Always true

Step-by-step explanation:

If you can't express the number as a ratio of integers, multiplying or dividing it by integers will not make it so you can.

If π is irrational, 2π is also irrational.

It is <em>always true</em> that the product of a rational and an irrational number is irrational.

7 0
3 years ago
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A polynomial f (x) has the
Helga [31]

a) The polynomial f(x) in <em>expanded</em> form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The <em>rational</em> function g(x) in <em>factored</em> form is g(x) = [(x - 6) · (x + 3)] / (x - 2). there is no <em>slant</em> asymptotes.

c) There is one <em>evitable</em> discontinuity at x = - 1, and one <em>definitive </em>discontinuity at x = 2, where there is a <em>vertical</em> asymptote.

<h3>How to analyze polynomial and rational functions</h3>

a) In the first part of this question we need to determine the equation of a polynomial in <em>expanded</em> form, derived from its <em>factor</em> form defined below:

f(x) = Π (x - rₐ), for a ∈ {1, 2, 3, 4, ..., n}         (1)

Where rₐ is the a-th root of the polynomial.

If we know that r₁ = 6, r₂ = - 1 and r₃ = - 3, then the polynomial in factor form is:

f(x) = (x - 6) · (x + 1) · (x + 3)

f(x) = (x - 6) · (x² + 4 · x + 4)

f(x) = (x - 6) · x² + (x - 6) · (4 · x) + (x - 6) · 4

f(x) = x³ - 6 · x² + 4 · x² - 24 · x + 4 · x - 24

f(x) = x³ + 10 · x² - 20 · x - 24

The polynomial f(x) in <em>expanded</em> form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The <em>rational</em> function is introduced below:

g(x) = (x³ + 10 · x² - 20 · x - 24) / (x² - x - 2)

g(x) = [(x - 6) · (x + 1) · (x + 3)] / [(x - 2) · (x + 1)]

g(x) = [(x - 6) · (x + 3)] / (x - 2)

The slope of the <em>slant</em> asymptote is:

m = lim [g(x) / x] for x → ± ∞

m = [(x - 6) · (x + 3)] / [x · (x - 2)]

m = 1

And the intercept of the <em>slant</em> asymptote is:

n = lim [g(x) - m · x] for x → ± ∞

n = Non-existent

Hence, there is no <em>slant</em> asymptotes.

c) There is <em>vertical</em> asymptote at a x-point if the denominator is equal to zero. There is one <em>evitable</em> discontinuity at x = - 1, and one <em>definitive </em>discontinuity at x = 2, where there is a <em>vertical</em> asymptote.

To learn more on asymptotes: brainly.com/question/4084552

#SPJ1

4 0
2 years ago
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