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lara [203]
3 years ago
9

Ted ran three miles in 2.8 minutes. At this rate, how long would it take him to run 33.3 miles?​

Mathematics
1 answer:
ohaa [14]3 years ago
4 0

Answer:

I believe it would take 30.96 minutes.

Step-by-step explanation:

2.8 divided by 3 = .93 miles per minute

.93 x 33.3 miles =30.969 minutes

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Match each set of vertices with the type of triangle they form.
Andrew [12]

Answer:  The calculations are done below.


Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

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Which of the following options have the same value as 25% of 64
tino4ka555 [31]

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Find the lateral area of the regular pyramid<br><br> L.A. =
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FOR REGULAR PYRAMID with those dimension.

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FOR HEXAGONAL PYRAMID with those dimension

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Step-by-step explanation:

Please the question asked for L.A of a REGULAR PYRAMID, but the figure is a HEXAGON PYRAMID.

Hence I solved for both:

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l = 8cm

L.A = 1/2 * 24 * 8

L.A = 1/2 ( 192)

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FOR AN HEXAGONAL PYRAMID

Lateral Area = 3a √ h^2 + (3a^2) / 4

Where:

a = Base Edge = 6

h = Height = 8

L.A = 3*6 √ 8^2 + ( 3*6^2) / 4

L.A = 18 √ 64 + ( 3 * 36) / 4

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L.A = 18 * 9.539

L.A = 171.71

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3 years ago
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