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Natalka [10]
2 years ago
15

**PLEASE HELP** What is the length of the altitude of the equilateral triangle below?

Mathematics
1 answer:
charle [14.2K]2 years ago
5 0
For any 30-60-90 triangle:
Let the length of hypotenuse = x
∴ The leg which faces the angle 30 = (1/2) x 
And the leg which faces the angle 60 = (√3/2) x 
So, The ratio between two legs will be 
= (1/2) x : (√3/2) x     ⇒⇒⇒⇒ divide over x
= 1/2  :  √3/2             ⇒⇒⇒⇒ multiply by 2
= 1  :  √3

Or the ratio will be

= (√3/2) x  :  (1/2) x    ⇒⇒⇒⇒ divide over x
= √3/2  :  1/2             ⇒⇒⇒⇒ multiply by 2
=  √3  :  1
==============================

The correct answer will be option B only

B.  1 : √3







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What is 252 = 11n + 10
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Answer:

n=22 or 252=11(22)+10

Step-by-step explanation: 252=11n+10. Start by attempting to isolate the varyable by subracting 10 from both sides--> 242=11n, divide by 11--> 22=n

PROOF 252=11*22+10=242+10=252


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If B = { p : p is a factor of 12 } list the elements of this set and find n ( B )​
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Answer:

B={1,2,3,4,6and12}

n (B) =6

Step-by-step explanation:

<h3>Greetings !</h3>

Factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The other factors of 12 are 1, 2, 4, and 12.

Hope it helps!

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2 years ago
Round to the nearest hundredth 0.207
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If you round it to the nearest hundredths the answer is .21
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3 years ago
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Is -18.75 a real number,irrational,rational,integer,whole or natural? it can be more than one choice
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3 years ago
the diameters of Douglas firs grown at a Christmas tree farm are normally distributed with a mean of 4 inches and a standard dev
aksik [14]

Answer:

Proportion of the trees will have diameters between 2 and 6 inches = 0.8164

Step-by-step explanation:

Given -

Mean (\nu )  = 4

Standard deviation (\sigma  ) = 1.5

Let X be the diameter of tree

proportion of the trees will have diameters between 2 and 6 inches =

P(2<  X<  6)   =  P(\frac{2 - 4 }{1.5}< \frac{X - \nu }{\sigma}<  \frac{6 - 4 }{1.5})

                         = P(\frac{-2 }{1.5}< Z<  \frac{2 }{1.5})     Put  [Z = \frac{X - \nu }{\sigma}]

                         =  P(-1.33< Z<  1.33)

                          = (Z<  1.33) - (Z<  -1.33)

                          = .9082 - .0918

                           = 0.8164

6 0
3 years ago
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