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yarga [219]
3 years ago
9

What is the perimeter of 36?

Mathematics
2 answers:
SOVA2 [1]3 years ago
3 0

Answer:

6

Step-by-step explanation:

sqrt 36 is 6*6 so is 6.

grandymaker [24]3 years ago
3 0
About 17.69 million miles
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Scrooge-
Neporo4naja [7]

Answer:

Step-by-step explanation:

8 0
3 years ago
If a circle is 9 pi cm what is the radius of the circle
olganol [36]

Answer:

<h2>          4.5 cm</h2>

Step-by-step explanation:

L = 2×π×r

9π cm = 2πr

9 cm = 2r

r = 4.5 cm

5 0
3 years ago
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
8.910312934 = n

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.
7 0
3 years ago
5. Point A is 3 units directly below (2, 2). What are the coordinates of point A? Answer: Point A (___)​
fiasKO [112]
A is (2, -1)

The X axis will not change since A is not going left or right of (2,2). You would count down 3 spaces for the y-axis and that would land you at -1 or you could solve by subtraction.

2-3= -1
3 0
1 year ago
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
juin [17]

Answer:

The Taylor series of f(x) around the point a, can be written as:

f(x) = f(a) + \frac{df}{dx}(a)*(x -a) + (1/2!)\frac{d^2f}{dx^2}(a)*(x - a)^2 + .....

Here we have:

f(x) = 4*cos(x)

a = 7*pi

then, let's calculate each part:

f(a) = 4*cos(7*pi) = -4

df/dx = -4*sin(x)

(df/dx)(a) = -4*sin(7*pi) = 0

(d^2f)/(dx^2) = -4*cos(x)

(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4

Here we already can see two things:

the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.

so we only will work with the even powers of the series:

f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....

So we can write it as:

f(x) = ∑fₙ

Such that the n-th term can written as:

fn = (-1)^{2n + 1}*4*(x - 7*pi)^{2n}

6 0
3 years ago
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