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lana66690 [7]
2 years ago
10

30 POINTS What is (f-g)(x) f(x)= x^4-3x^2+5x-7 g(x)= x^3-3x^2+3x-1

Mathematics
2 answers:
AVprozaik [17]2 years ago
6 0

Answer:

x^4 - x^3 + 2x -6

Step-by-step explanation:

(f-g)(x) = x^4-3x^2+5x-7 - ( x^3-3x^2+3x-1 ) = x^4 - 3x^2+5x-7 - x^3 + 3x^2 - 3x + 1  = x^4 - x^3 + 2x -6

kobusy [5.1K]2 years ago
5 0

Step-by-step explanation:

x^4-3x^2+5x-7 – x^3-3x^2+3x-1

x^4-3x^2+5x-7 – x^3+3x^2-3x+1

x^4-3x^2+3x^2-3x +5x -x^3 -7+1

x^4+2x -x^3 -6

x^4-x^3+2x-6

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The height of a punted football can be modeled with the quadratic function 0.01x^2 + 1.18x+2.
Fantom [35]

The function of the path of the punted ball is a quadratic function which

follows the path of a parabola.

The correct responses are;

Part A: The coordinates of the vertex is \underline {(59, \, 36.81)}

Part B: The maximum height of the punt is <u>36.81 ft.</u>

Part C: The defensive player must reach up to <u>7.65 feet</u> to block the punt.

Part D: The distance down the field the ball will go without being blocked is approximately <u>119.67 ft.</u>

<u />

Reasons:

The function for the height of the punted ball is; h = -0.01·x² + 1.18·x + 2

Assumption; The distances are feet.

Part A: By completing the square, we have;

f(x) = -0.01·x² + 1.18·x + 2

100·f(x) = -x² + 118·x  + 200

-100·f(x) = x² - 118·x  - 200

x² - 118·x + (118/2)²= 200 + (118/2)²

(x - 59)² = 200 + (59)² = 3681

(x - 59)² - 3681

At the vertex, -3281 = -100·f(x)

∴ f(x) at the vertex = -3681/-100 = 36.81

\mathrm{\underline{Coordinate \ of \ the \ vertex = (59, \, 36.81)}}

Part B: The maximum height is given by the y-value at the vertex = 36.81 ft.

Part C: When <em>x</em> = 5, we have;

h = -0.01·x² + 1.18·x + 2

h = -0.01 × 5² + 1.18 × 5 + 2 = 7.65

The defensive player must reach up to 7.65 feet to block the punt

Part D: The distance the ball will go before it hits the ground is given by

the function, for the height as follows;

h = -0.01·x² + 1.18·x + 2 = 0

From the completing the square method, above, we get;

-0.01·x² + 1.18·x + 2 = 0

x² - 118·x  - 200 = 0

x² - 118·x + (118/2)²= 200 + (118/2)²

x² - 118·x + (59)²= 200 + (59)² = 3681

(x - 59)² = 3681

x - 59 = ±√3681

x = 59 ± √3681

x = 59 + √3681 ≈ 119.67

The distance down the field the ball will go without being blocked, x ≈ <u>119.67 ft.</u>

<u />

Learn more here:

brainly.com/question/24136952

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2 years ago
Which property is used in the statement? <br> If z=3+1, then 3+1=z
saul85 [17]
This is the symmetric property of equality, and in a more general form it can be written  as: if a=b then b=a.

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3 years ago
Use lengths |OQ| = 9.4 units and |OQ’| = 17.1 units to determine the scale factor r of dilation D (round to the nearest hundredt
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Answer:

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Step-by-step explanation:

We have been given that side length OQ of our prei-mage is 9.4 units and side length OQ' after dilation is 17.1 units. Since our pre-image (original image) is smaller than our new image, so our scale factor (r) will be greater than 1.

Since we know that in a dilation, the sides of the pre-image and the corresponding sides of the image are proportional, so we will use proportion to find a scale factor our given side lengths as:

\frac{OQ'}{OQ}=\frac{17.1}{9.4}

\frac{OQ'}{OQ}=1.8191

Upon multiplying both sides of our equation by OQ we will get,

\frac{OQ'}{OQ}\times OQ=1.8191\times OQ

OQ'=1.8191\times OQ

Upon rounding our answer to nearest hundredths place we will get,

OQ'\approx 1.82\times OQ

Since side length OQ' is 1.82 times side length OQ, therefore, our scale factor (r) will be 1.82 units.

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