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Sergio039 [100]
3 years ago
12

How many multiples of 5 are there between 199 and 1,198? Hint: an = a1 + d(n − 1), where a1 is the first term and d is the commo

n difference.
Mathematics
1 answer:
GaryK [48]3 years ago
4 0
An=a1+d (n-1)
A1=200 since that's the first term that can be a multiple of 5
N=? That's what we need to find
An=1195 since that's the last multiple of 5 that we can use
D=5
Plug in
1195=200+5 (n-1)
-200 both sides
995=5 (n-1)
Distribute 5 to n-1
995=5n-5
+5 both sides
1000=5n
÷5 both sides
N=200 there are 200 multiples of 5 in between 199 and 1198
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X = 2 or X = -10

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The height, h, of the equilateral triangle is given by h =5cotΘ, where Θ is 30 degrees. Can someone help me with this? I don't u
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You don't even need the picture to solve this one.

You said that      h = 5 cot(Θ) ,    and you said that  Θ  is 30 degrees.
All you need now is to find the cotangent of  Θ, plop that into the equation,
and the solution practically jumps off the paper into your lap.

To find the cotangent of 30 degrees, you can use a calculator, look it up
in a book, read it off of a slide rule if you have one, draw a picture of a
30-60-90 right triangle etc.  You'll find that the cotangent of 30 degrees
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So your equation is    h = 5 (1.732)  =  <em>8.66  </em>(rounded)

Apparently, somebody gave you the equation, and asked you to find 'h'.
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