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Nostrana [21]
3 years ago
13

Which of the following illustrates the truth value of the given conditional statement?

Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0
<h2>B . TT →T</h2><h2>..............</h2><h2>................</h2><h2>....................</h2>
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A pilot can travel for 140 miles with the wind in the same amount of time as 360 miles against the wind. Find the speed of the w
quester [9]

Answer:

  -88 mph

Step-by-step explanation:

We can use the relation ...

  time = distance/speed

to compare the times in the two directions.

  \dfrac{140}{200+w}=\dfrac{360}{200-w}\\\\140(200-w)=360(200+w)\\\\200(140-360)=w(360+140)\\\\w=\dfrac{200(-220)}{500}=-88 \quad\text{miles per hour}

The wind speed is -88 miles per hour.

_____

The problem statement tells us the travel is slower <em>with</em> the wind than <em>against</em> the wind. Hence "with the wind" must be subtracting from the net speed. That is, the wind speed is negative.

4 0
3 years ago
Please answer! explain the last one
masha68 [24]

If Nick starts with 20, we can just plug it into p/2 - 12 to get 10-12 or -2. This means Nick did not get 20 cards. Now we have answered both. Two-liner!

8 0
3 years ago
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

5 0
2 years ago
C=5(350-32)/9 in celsius pls help
nevsk [136]
C= 5(318)/9
= 1590/9
= 176.67 degrees celsius
8 0
3 years ago
Answer 69 all of the parts pleaseeee
marin [14]
Part b is square. a is paralel iogram is 1 2 3 5. 
part c is yes the are both parralelograms because the both a at east one set of parrell lines

6 0
3 years ago
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