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bezimeni [28]
3 years ago
8

1) What radius of a circle is required to inscribe an equilateral triangle with an area of 15.588 in2 and an altitude of 5.196 i

n? (round to nearest tenth)

Mathematics
2 answers:
Andrej [43]3 years ago
8 0
We know that
 the distance from the centroid of the triangle to one of the vertices is the radius of the circle <span>required to inscribe an equilateral triangle.

[distance </span>centroid of the triangle to one of the vertices]=(2/3)*h
h=the <span>altitude  of the equilateral triangle-----> 5.196 in
so
</span>[distance centroid of the triangle to one of the vertices]=(2/3)*5.196
[distance centroid of the triangle to one of the vertices]=3.464 in----> 3.5 in

the radius is equal to the distance of the centroid of the triangle to one of the vertices
hence
the radius is 3.5 in

the answer is
the radius is 3.5 in

slamgirl [31]3 years ago
6 0

It's 3.5, 100% sure.



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Determine the astm grain size number if 33 grains per square inch are measured at a magnification of 270×
strojnjashka [21]
8.9

The equation for the grain size is expressed as the equality:
Nm(M/100)^2 = 2^(n-1)
where
Nm = number of grains per square inch at magnification M.
M = Magnification
n = ASTM grain size number

Let's solve for n, then substitute the known values and calculate.
Nm(M/100)^2 = 2^(n-1)

log(Nm(M/100)^2) = log(2^(n-1))
log(Nm) + 2*log(M/100) = (n-1) * log(2)
(log(Nm) + 2*log(M/100))/log(2) = n-1
(log(Nm) + 2*log(M/100))/log(2) + 1 = n

(log(33) + 2*log(270/100))/log(2) + 1 = n
(1.51851394 + 2*0.431363764)/0.301029996 + 1 = n
(1.51851394 + 0.862727528)/0.301029996 + 1 = n
2.381241468/0.301029996 + 1 = n
7.910312934 + 1 = n
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So the ASTM grain size number is 8.9

If you want to calculate the number of grains per square inch, you'd use the
same formula with M equal to 1. So:
Nm(M/100)^2 = 2^(n-1)
Nm(1/100)^2 = 2^(8.9-1)
Nm(1/10000) = 2^7.9
Nm(1/10000) = 238.8564458
Nm = 2388564.458

Or about 2,400,000 grains per square inch.
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serg [7]

Answer:

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