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V125BC [204]
3 years ago
15

An integer is three more than another integer. Twice the larger integer is two less

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
3 0

Answer:

The integers are 4 and 7 or -2 and 1.

Step-by-step explanation:

You can make a system of equations with the description of the two integers.

1. x = y + 3

2. 2x + 2 = y^2

The simplest and the fastest way to solve this system in this case is substitution. You can substitute x for y + 3 in the second equation.

1. x = y + 3

2. 2(y + 3) + 2 = y^2

Now simplify and solve the second one. For convenience, I will just disregard the first equation for now.

2y + 6 + 2 = y^2

y^2 - 2y - 8 = 0

You can factor this equation to solve for y.

(y - 4) (y + 2) = 0

y = 4, y = -2

Now we can substitute the value of y for x in the first equation.

x = 7, x = 1

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The value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

<h3>What is the value of f[-4] and g°f[-2]?</h3>

Given the function;

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For f[ -4 ], we substitute -4 for every variable x in the function.

f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

g(\frac{3x-2}{x+1}) =  (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) =  \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) =  13

Therefore, the value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

Learn more about composite functions here: brainly.com/question/20379727

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The given curve has equation

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and its derivative is

d<em>r</em>/d<em>θ</em> = -8 sin(<em>θ</em>)

When <em>θ</em> = <em>π</em>/3, we have <em>r</em> (<em>π</em>/3) = 13, and d<em>r</em>/d<em>θ</em> (<em>π</em>/3) = -4√3.

Differentiate these with respect to <em>θ</em> :

d<em>y</em>/d<em>θ</em> = d<em>r</em>/d<em>θ</em> sin(<em>θ</em>) + <em>r(θ)</em> cos(<em>θ</em>)

d<em>x</em>/d<em>θ</em> = d<em>r</em>/d<em>θ</em> cos(<em>θ</em>) - <em>r(θ</em>) sin(<em>θ</em>)

In polar coordinates, we have

<em>y(θ)</em> = <em>r(θ)</em> sin(<em>θ</em>)

<em>x(θ)</em> = <em>r(θ)</em> cos(<em>θ</em>)

and when <em>θ</em> = <em>π</em>/3, we have <em>y</em> (<em>π</em>/3) = 13√3/2 and <em>x</em> (<em>π</em>/3) = 13/2.

The slope of the tangent line to the curve is d<em>y</em>/d<em>x</em>. By the chain rule,

d<em>y</em>/d<em>x</em> = d<em>y</em>/d<em>θ</em> • d<em>θ</em>/d<em>x</em> = (d<em>y</em>/d<em>θ</em>) / (d<em>x</em>/d<em>θ</em>)

d<em>y</em>/d<em>x</em> = (d<em>r</em>/d<em>θ</em> sin(<em>θ</em>) + <em>r(θ)</em> cos(<em>θ</em>)) / (d<em>r</em>/d<em>θ</em> cos(<em>θ</em>) - <em>r(θ</em>) sin(<em>θ</em>))

When <em>θ</em> = <em>π</em>/3, the slope is

d<em>y</em>/d<em>x</em> = (-4√3 sin(<em>π</em>/3) + 13 cos(<em>π</em>/3)) / (-4√3 cos(<em>π</em>/3) - 13 sin(<em>π</em>/3))

d<em>y</em>/d<em>x</em> = (-4√3 (√3/2) + 13 (1/2)) / (-4√3 (1/2) - 13 (√3/2))

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