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3 years ago

If a circle with a diameter of 4 inches rotates 1000°, what distance will any one spot on

Mathematics

1 answer:

Orlov [11]3 years ago

5 0

We first convert from degrees to radians multiplying by π/180:

θ = 1000° × π/180 = 17.45 rad

The radius is:

r = d/2 = 4 in /2 = 2 inch

So, The distance traveled is given by the product of the angle in radians and the radius. This is:

d = θr = (17.45 rad)(2 inch) = 34.9 inch

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Find f(2) if f(x) = (x 1)2.

Katyanochek1 [597]

I hope this helps you

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f (2)=(2+1)^2=3^2=9

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4 years ago

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There are two roots because of the....

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Need to identify + and - roots of the radicand.

Step-by-step explanation:

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3 years ago

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Abc is an equilateral triangle and has an area of 8/5. point d is the midpoint of side ab, point e is the midpoint of side bc, a

nlexa [21]

Hello,

There are 4 congurent equilateral triangles in ABC and each has an area of (8/5)/4=2/5.
To form the parallelogram we need 2 triangles: 2* 4/5=4/5.

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3 years ago

Help question on the image.

Simora [160]

Answer:

Step-by-step explanation:

We want to compare the pot holder hole dimension to the pot dimension. The hole diameter is the overall pot holder diameter less the wall thickness on either side:

hole diameter = (20 1/5) -2(1 2/7) = 20 -2(1) +1/5 -(2)(2/7)

= 18 +(1/5 -4/7) = 18 +(7 -20)/35 = 18 -13/35

= 17 22/35 . . . . cm

We want to compare this to 17 4/5. We can do that by using a common denominator for the fractions.

17 4/5 = 17 + (4/5)(7/7) = 17 28/35 . . . . cm

The pot has a diameter of 17 28/35 cm; the pot holder has a hole diameter of 17 22/35 cm, so the hole is smaller than the pot.

The pot will not fit into the plant pot holder.

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You can use your calculator to find the free space around the pot by computing ...

20 1/5 -2(1 2/7) -17 4/5 = free space = -6/35 . . . . cm

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2 years ago

Please Help!!!!!! So i know how to reflect over line y=-x but how would i do it over line y=-x+6. An example would be helpful an

Westkost [7]

To reflect in the line y = -x + 6, we translate everything down 6 first. This will make it seem like we are reflecting in the line y = -x
                                 (x,y) → (x, y-6)
Then, to reflect in the line y = -x, we switch the x- and y-coordinates and then make them negative
                                 (x,y) → (-(y-6), -x)
Then we move everything back up again
                                 (x,y) → (-(y-6), -x + 6)
I will present to you an example. Reflect the point (-4, 8) in the line y = -x + 6.
                                 (-4,8) → (-(8-6), -(-4) + 6) → (-2, 10)
You should graph this out to confirm with the reflection line.

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4 years ago