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Len [333]
3 years ago
8

Please help with this question

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
8 0

Answer:

the cost per mile :)


Maru [420]3 years ago
4 0
She's right. oh and do you do k12
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Vesnalui [34]

Step-by-step explanation:

csc x / (cot x + tan x)

Write in terms of sine and cosine.

(1 / sin x) / [(cos x / sin x) + (sin x / cos x)]

Multiply top and bottom by sin x.

1 / [cos x + (sin²x / cos x)]

Multiply top and bottom by cos x.

cos x / (cos²x + sin²x)

Use Pythagorean identity.

cos x / 1

cos x

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Nookie1986 [14]
65-14=65-5-x
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Explanation=
65-14=65-5-x
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6 0
3 years ago
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Annette [7]
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Which of these four sets of side lengths will form a right triangle? Set 1 4 cm, 5 cm, 6 cm Set 2 8 in., 12 in., 20 in. Set 3 10
Alenkasestr [34]

Answer:  The correct option is (D) SET 4.

Step-by-step explanation:  We are to select the correct set of side lengths that will form a right-angled triangle.

To form a right-angled triangle, we must have the following relation:

<em>Perpendicular² + Base² = Hypotenuse².</em>

<em>Hypotenuse is the length of the largest side; perpendicular and base are the two legs of the triangle.</em>

SET 1 :  14 cm, 5 cm, 6 cm.  

We have

5^2+6^2=25+36=61,\\\\14^2=196.

Therefore,

<em>Perpendicular² + Base² ≠ Hypotenuse².</em>

So, this set will not form a right-angled triangle.

SET 2 :  8 in., 12 in., 20 in.  

We have

8^2+12^2=64+144=208,\\\\20^2=400.

Therefore,

<em>Perpendicular² + Base² ≠ Hypotenuse².</em>

So, this set will not form a right-angled triangle.

SET 3 :  10 mm, 20 mm, 30 mm.  

We have

10^2+20^2=100+400=500,\\\\30^2=900.

Therefore,

<em>Perpendicular² + Base² ≠ Hypotenuse².</em>

So, this set will not form a right-angled triangle.

SET 4 :  12 ft, 16 ft, 20 ft.  

We have

12^2+16^2=144+256=400,\\\\20^2=400.

Therefore,

<em>Perpendicular² + Base² = Hypotenuse².</em>

So, this set will form a right-angled triangle.

Thus, the SET 4 will form a right-angles triangle.

Option (D) is correct.

5 0
3 years ago
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just olya [345]
They sold an average of 79 pies a day
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