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Anit [1.1K]
3 years ago
6

Please help me with math pt.1

Mathematics
2 answers:
lesya [120]3 years ago
6 0

Answer:

No. of miles Jake rode today = 6.6×2.4 times

= 15.84 miles

1 gallon of gas = 27.6 miles

7 gallons of gas = 193.2 miles

Myla's car can go 193.2 miles with 7 gallons of gas.

All expressions except 7.33×6.

kvv77 [185]3 years ago
3 0

Answer:

i don't think you're in high school

#1. 15.84 (6.6 * 2.4)

#2 193.2 (27.6 * 7)

#3 B and E

Step-by-step explanation:

6.6 * 2.4

27.6 * 7

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Collin bought 1/2 a pound of chocolate at Rocky Mountain Chocolate factory. Later, they went to The Sweet Shoppe and he bought 6
Lera25 [3.4K]

Answer:

1 1/6

Step-by-step explanation:

6/9 times 2/2=12/18

1/2 times 9/9=9/18

12/18+9/18=21/18

21/18=1 3/18

1 3/18=1 1/6

6 0
3 years ago
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Simplify. 4^3 x 4^4/ 4^9<br><br> A. 4^16<br><br> B. 4^2<br><br> C. 1/4^2<br><br> D 1/4
forsale [732]
The answer is A. 4^16
3 0
3 years ago
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The local bakery uses 1.75 cups of flour in each batch of cookies. The bakery used 5.25 cups of flour this morning. If there are
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Answer: 180 cookies


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5 0
3 years ago
This is the call plan for Robert's cell phone:
butalik [34]
To answer this, first make an equation representing the data: 

15+ 0.20X= Bill

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4 0
3 years ago
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Linda goes water-skiing one sunny afternoon. After skiing for 15 min, she signals to the driver of the boat to take her back to
trasher [3.6K]
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2


abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.


d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
4 0
3 years ago
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