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sineoko [7]
3 years ago
11

Name of devices is used for drawing curves which can't be drawn by a compass​

Mathematics
1 answer:
Snowcat [4.5K]3 years ago
6 0
A protractor can be used!
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What is 1/6 divided by 1/2
Blababa [14]

Answer:

The answer is 3

Step-by-step explanation:

I put the answer simplified if needed.

8 0
2 years ago
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G = {(5,3), (2, 3), (6,4)} Is G^-1 a function and why?
sleet_krkn [62]

Answer:

The inverse relation G^(-1) is not a function. Why not? Because the y value y = 3 is paired up with more than one x value (x = 5, x = 2). The inverse relation G^(-1) is the set shown below

{(3,5), (3,2), (4,6)}

All I've done is swap the (x,y) values for each ordered pair to form the inverse relation. As you can see, x = 3 leads to multiple y value outputs which is why this relation is not a function. So in short, the answer is choice C. To have the inverse relation be a function, each value in the original domain must map to exactly one value in the range only. However that doesn't happen as the domain values map to an overlapping y value (y = 3).

5 0
3 years ago
A 17.2 gallon gasoline tank is 1/2 full. How many gallons will it take to fill the tank?
love history [14]

Answer:

8.6

Step-by-step explanation:

We know that the tank is filled halfway, and it can hold 17.2 gallons, so we can just divide to figure out how much is half the tank (the amount we need to fill it)

17.2 / 2 = 8.6

We can check our work by multiplying

8.6 x 2 = 17.2

So it would take 8.6gal to fill it up. I hope this helps :)

3 0
3 years ago
25 POINTS WORTH
Vikki [24]

Answer:

where is the picture?

Step-by-step explanation:

8 0
3 years ago
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The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
2 years ago
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