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3 years ago

Help asap!!!!! im in a quiz rn-

Mathematics

1 answer:

a_sh-v [17]3 years ago

3 0

Answer:

C. 31.4 ft

Step-by-step explanation:

Given

Diameter (d) = 10 ft

Now

Circumference

= πd

= 3.14 * 10

= 31.4 ft

Hope it will help :)❤

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Illusion [34]

Answer:

Option (c) is correct

Step-by-step explanation:

Case (a)

A = 3 + 5i = (3, 5)

B = 2 + 2i = (2, 2)

C = 5i = (0, 5)

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$AB = \sqrt{(2-3)^{2}+(2-5)^{2}}=\sqrt{10}$

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For the triangle to be right angles triangle

$BC^{2}=AB^{2}+CA^{2}$

Here, it is not valid, so these are not the points of a right angled triangle.

Case (b)

A = 2i = (0, 2)

B = 3 + 5i = (3, 5)

C = 4 + i = (4, 1)

Use the distance formula to find the distance between two points

$AB = \sqrt{(3-0)^{2}+(5-2)^{2}}=\sqrt{18}$

$BC = \sqrt{(4-3)^{2}+(1-5)^{2}}=\sqrt{17}$

$CA = \sqrt{(4-0)^{2}+(1-2)^{2}}=\sqrt{17}$

For the triangle to be right angles triangle

$AB^{2}=BC^{2}+CA^{2}$

Here, it is not valid, so these are not the points of a right angled triangle.

Case (c)

A = 6 + 4i = (6, 4)

B = 7 + 5i = (7, 5)

C = 8 + 4i = (8, 4)

Use the distance formula to find the distance between two points

$AB = \sqrt{(7-6)^{2}+(5-4)^{2}}=\sqrt{2}$

$BC = \sqrt{(8-7)^{2}+(4-5)^{2}}=\sqrt{2}$

$CA = \sqrt{(8-6)^{2}+(4-4)^{2}}=\sqrt{4}$

For the triangle to be right angles triangle

$CA^{2}=BC^{2}+AB^{2}$

Here, it is valid, so these are the points of a right angled triangle.

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