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Oksanka [162]
3 years ago
5

70 pointsss for the correct answer

Mathematics
1 answer:
nignag [31]3 years ago
3 0

Answer:

f(x) = 12x³ - 19x² + 17x - 17

Step-by-step explanation:

We can the numerator by multiplying the denominator by the quotient, and then adding the remainder.

We're given a denominator of 4x - 5

a quotient of 3x² - x + 3

and a remainder of -2

so the original numerator is:

(4x - 5)(3x² - x + 3) - 2

= 12x³ - 4x² + 12x - 15x² + 5x - 15 - 2

= 12x³ - 19x² + 17x - 17

Let's test that by dividing it by 4x -5 with long division:

\ \ \ \ \ \ \ \ \ \  3x^2 - x + 3\\4x - 5\overline{)12x^3 - 19x^2 + 17x - 17}\\.\  \ \ \ \ \ \ \ \ \underline{2x^3 - 15x^2}\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \  -4x^2 + 17x\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{-4x^2 + 5x} \\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 12x - 17\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{12x - 15}\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2R

That matches, so we know our answer's right, and:

f(x) = 12x³ - 19x² + 17x - 17

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Crickets can jump with a vertical velocity of up to 14 ft/s. Which equation models the height of such a jump, in feet, after t s
Y_Kistochka [10]

Answer:

h = 147 - 16t^2.

Step-by-step explanation:

Use the following equation of motion

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So here we have:

h = 14t + 1/2 * -32 * t^2

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A ball is thrown into the air with a velocity of 34 ft/s. It's height in feet after t seconds is given by y=34t-26(t)^2. Find th
kkurt [141]

Answer:

The average velocity of the ball at the given time interval is -122.3 ft/s

Step-by-step explanation:

Given;

velocity of the ball, v = 34 ft/s

height of the ball, y = 34t - 26t²

initial time, t₀ = 3 seconds

final time, t = 3 + 0.01 = 3.01 seconds

At t = 3 s

y(3) = 34(3) - 26(3)² = -132

The average velocity of the ball in ft/s is given as;

V_{avg} = \frac{y(3.01)-y(3)}{3.01 -3}\\\\V_{avg} = \frac{34(3.01)-26(3.01)^2-y(3)}{3.01 -3}\\\\V_{avg} = \frac{-133.223-y(3)}{0.01}\\\\V_{avg} = \frac{-133.223-(-132)}{0.01}\\\\V_{avg} =\frac{-1.223}{0.01}\\\\V_{avg} = -122.3 \ ft/s

Therefore, the average velocity of the ball at the given time interval is -122.3 ft/s

8 0
3 years ago
Find the length and width of a rectangle that has the given area and a minimum perimeter. Area: 162 square feet
Gnom [1K]

Answer:

The width and length of rectangle is 12.728 m

Step-by-step explanation:

Let the length of the rectangle = L

let the width of the rectangle = W

The subjective function is given by;

F(p) = 2(L + W)

F = 2L + 2W

Area of the rectangle is given by;

A = LW

LW = 162 ft²

L = 162 / W

Substitute in the value of L into subjective function;

f = 2l + 2w\\\\f = 2(\frac{162}{w} )+2w\\\\f = \frac{324}{w} + 2w\\\\\frac{df}{dw} = \frac{-324}{w^2} +2\\\\

Take the second derivative of the function, to check if it will given a minimum perimeter

\frac{d^2f}{dw^2}= \frac{648}{w^3} \\\\Thus, \frac{d^2f}{dw^2}>0, \ since,\frac{648}{w^3} >0 \ (minimum \ function \ verified)

Determine the critical points of the first derivative;

df/dw = 0

\frac{-324}{w^2} +2 = 0\\\\-324 + 2w^2=0\\\\2w^2 = 324\\\\w^2 = \frac{324}{2} \\\\w^2 = 162\\\\w= \sqrt{162}\\\\w = 12.728 \ m

L = 162 / 12.728

L = 12.728 m

Therefore, the width and length of rectangle is 12.728 m

3 0
3 years ago
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