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Dmitry [639]
2 years ago
10

The table shows the total amount of snowfall for different numbers of hours in Alphaville. The graph shows the total amount of s

nowfall for different numbers of hours in Beta City. Which statement is true?
A table with 2 rows, Hours, x, and Snowfall, in inches, y. Column 1 says 2, 2 point 5. Column 2 says 3, 3 point 7 5. Column 3 says 5, 6point 2 5. Column 4 says 6, 7 point 5. The first quadrant of a coordinate plane with Number of Hours on the x axis and Snowfall, in inches, on the y axis. A straight line is graphed. The line passes through the points zero, zero and 8, twelve.
A
It snowed 1.25 inches per hour in Alphaville, which is the greater rate of snowfall.

B
It snowed 1.25 inches per hour in Beta City, which is the lesser rate of snowfall.

C
It snowed 1.5 inches per hour in Beta City, which is the greater rate of snowfall.

D
It snowed 1.5 inches per hour in Alphaville, which is the lesser rate of snowfall.
Mathematics
2 answers:
ikadub [295]2 years ago
5 0

Answer:

B

Step-by-step explanation:

nadezda [96]2 years ago
5 0

Answer:C

Step-by-step explanation:

The correct answer is C. It snowed one and a quarter inches per hour in Alphaville. It snowed one and a half inches per hour in Beta City, which is the greater rate of snowfall.

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Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

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Vertical distance

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Divide each side of (1) by 600cosθ.

(3) \, t =\dfrac{2}{\cos \theta}

Substitute (3) into (2)

-50 = 600t \sin \theta - 4.9t^{2} =  600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )

Set up a quadratic equation

\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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3 years ago
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