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Aleonysh [2.5K]

2 years ago

13

On a map Rossville and Smithtown are 5 1/2 inches apart. If the actual distance between the town is 16 1/2 miles, to what unit r

ate is the map drawn?
A. 1 inch = 2 1/2 miles
B. 1 inch = 3 miles
C. 1 inch = 3 1/3 miles
D. 1 inch = 11 miles

2 answers:

nikdorinn [45]2 years ago

6 0

The answer is C because it obviously is 1 2 3 4 5 6 7 8 9 10

melisa1 [442]2 years ago

5 0

Answer:

c

Step-by-step explanation:

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A jogger goes half a mile north and then turns west. If the jogger finishes 1.3 miles away from the starting point, how far west

Ratling [72]

In order to solve this problem, you must draw a right triangle first to help you visualize the path of the jogger.
You are given 0.5 miles north (which will be the height of the right triangle), and 1.3 miles (which will be the hypotenuse of the triangle).
Using Pythagorean Theorem: a²+b²=c², we have:
0.5² + b² = 1.3²
b = 1.3²-0.5²
b = 1.2 miles

3 0

2 years ago

Brainliest Award +100 Points for the first correct answer.

katen-ka-za [31]

Answer:

min = 64.2m ; max = 86.2m

Step-by-step explanation:

$38.25m \leq AC < 38.35m
\\ 11.5deg \leq B < 12.5deg
\\ 19.5deg \leq D < 20.5deg$

$d_{min} = BC_{LB} - CD_{UB}
\\ \\ d_{min} = \frac{AC_{LB}}{tan(B)_{UB}} - \frac{AC_{UB}}{tan(D)_{LB}}
\\ \\ d_{min} = \frac{38.25}{tan(12.5)} - \frac{38.35}{tan(19.5)}
\\ \\ d_{min} = 64.2m (3sf)$

$d_{max} = BC_{UB} - CD_{LB}
\\ \\ d_{max} = \frac{AC_{UB}}{tan(B)_{LB}} - \frac{AC_{LB}}{tan(D)_{UB}}
\\ \\ d_{max} = \frac{38.35}{tan(11.5)} - \frac{38.25}{tan(20.5)}
\\ \\ d_{max} = 86.2m (3sf)$

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2 years ago

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1 year ago

WILL GIVE BRAINLIEST!<br> PLEASE SOLVE

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19.
x=4
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20.
x=10.75
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3 years ago

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Brilliant_brown [7]

180- (80-10x)

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2 years ago