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kirill115 [55]
3 years ago
8

If r is the radius of a circle and d is its diameter, which of the following is an

Mathematics
1 answer:
solong [7]3 years ago
5 0

Answer:

A

Step-by-step explanation:

because r is 1/2 of d so you would have 1 d instead of 2 r

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The cost of a movie ticket is increased by 15%. The old price was five dollars how much are they now?
Grace [21]

Answer:

5.75 is the answer.

$5.00 x 0.15=0.75

$5.00+0.75=$5.75

4 0
3 years ago
Gary is using an indirect method to prove that segment DE is not parallel to segment BC in the triangle ABC shown below:
aliya0001 [1]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Based on the situation above the inequality will he use to contradict the assumption is <span>
4:10 ≠ 6:14</span>
if DE is parallel to BC
then
4: (4+5) = 6 : (6 + 8)
4 0
3 years ago
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Laura is 6 feet tall how tall is she in inches
daser333 [38]
Laura is 72 inches tall. 
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3 years ago
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In a regular triangle ABC with side 1, two squares MNKL, RKPT are drawn such that points M, L, R are on the side AC (the order o
Orlov [11]

Answer:

  • MN = (21 -6√3)/37 ≈ 0.286694
  • RK = (14√3 -12)/37 ≈ 0.331046

Step-by-step explanation:

In the attached figure, we have defined LM to be length x. Then the other lengths on side AC are ...

  AM = LR = x/√3

  RC = (2/√3)RK = (2/√3)(2/√3)x = 4/3x

Then the sum of lengths along AC is ...

  AC = AM +ML +LR +RC

  1 = x(1/√3 +1 +1/√3 +4/3) = x(7/3 +2/√3) = x(7√3 +6)/(3√3)

Then the value of x is ...

  x=\dfrac{3\sqrt{3}}{7\sqrt{3}+6}=\dfrac{3\sqrt{3}(7\sqrt{3}-6)}{(7\sqrt{3})^2-6^2}=\dfrac{3(21-6\sqrt{3})}{3(49-12)}\\\\\boxed{MN=\dfrac{21-6\sqrt{3}}{37}}\\\\RK=\dfrac{2\sqrt{3}}{3}MN\\\\\boxed{RK=\dfrac{14\sqrt{3}-12}{37}}

8 0
3 years ago
The shaded prism below is created from the rectangular box as shown. Points A, B, and C are midpoints of their respective edges.
spayn [35]

Answer:

The volume of the prism is 1/4 of the volume of the rectangular box.

Step-by-step explanation:

The figure alluding to the exercise is required, according to the description I will attach the one that must be to be able to solve the exercise.

The first thing is that the cross sections of the prism are triangles and in addition those triangles are congruent to each other with areas equal to the area of ​​the base triangle.

By congruence we can say that the triangle has 1/4 of the area of ​​the base rectangle. We can affirm that the height of the prism is equal to the height of the rectangular box.

Now, the Cavalieri principle states that if two solids have the same height and their cross-sectional areas taken parallel and at equal distances from their bases are always equal, then they have the same volume.

now in this case the cross-sectional areas (parallel to the base) of the prism and the cross-sectional areas (parallel to the base) of the cuboid with a height equal to that of the rectangular box and the length, width of half of the sizes of the rectangular box are always the same.

Which means that the volume of the parallelepiped is 1/4 the volume of the rectangular box and thanks to this we can say that the volume of the prism is 1/4 of the volume of the rectangular box.

5 0
4 years ago
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