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shusha [124]
2 years ago
13

COVERT 28,957 ft = mi ft.

Mathematics
2 answers:
Wittaler [7]2 years ago
7 0
28,957 is equal to 5 miles, 2,557 ft
cluponka [151]2 years ago
3 0

Answer:

5.4843 miles

Step-by-step explanation:

28957 ft= 5.4843 miles

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Sonoma bikes 5 miles to Paige's house. On a map, they measure that distance as 5/6cm. The same map shows that the mall is 3 1/2c
Elden [556K]
Simplify fraction then subtract fractions
7 0
3 years ago
Passing through the coordinate (10,12) parallel to y=6
erastova [34]

Answer:

y = 12

Step-by-step explanation:

A line parallel to y = 6 is, like y = 6, horizontal.  Such a line passing through (10,12) is y = 12.


7 0
3 years ago
Danielle and David each have a job. Danielle earns $15 an hour. She also receives a $50 weekly bonus. Danielle wrote an equation
S_A_V [24]
For starters, create an equation to show David's earnings. We can do this using Danielle's as a basis, which is set up as y=(# of hours)x+(bonus). This gives us y=12x+80. Now, as we need both their ys to be equal, we just set both equations equal to each other, making 15x+50=12x+80. Now, we solve for x, starting with 15x+50=12x+80, subtracting 12x from both sides to get 3x+50=80, subtracting 50 from both sides to get 3x=30, and dividing three from both sides to get x=10. To check, we just plug in our answer to both equations and see if the ys match up. With Danielle's equation, we get y=15(10)+50=150+50=200 and with David's equation, we get y=12(10)+80=120+80=200, proving that our answer is correct.
5 0
3 years ago
Read 2 more answers
Can somebody solve this
Gemiola [76]
The answer is 36
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4 0
2 years ago
Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En
ArbitrLikvidat [17]

Answer:

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, c) x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right), y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right).

Step-by-step explanation:

The equation of the circle is:

x^{2} + (y-1)^{2} = 16

After some algebraic and trigonometric handling:

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t

Where:

\frac{x}{4} = \cos t

\frac{y-1}{4} = \sin t

Finally,

x = 4\cdot \cos t

y = 1 + 4\cdot \sin t

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

c) x = 4\cdot \cos t'', y = 1 + 4\cdot \sin t''

Where:

4\cdot \cos t' = 0

1 + 4\cdot \sin t' = 5

The solution is t' = \frac{\pi}{2}

The parametric equations are:

x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right)

y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)

7 0
3 years ago
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