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Colt1911 [192]
3 years ago
6

Write a pair of fractions as a pair of fractions with a common denominator using the fractions 1/2 and 1/3.

Mathematics
1 answer:
Vinvika [58]3 years ago
4 0

Answer:

3/6, 2/6.

Step-by-step explanation:

The lowest common denominator of 2 and 3 is 6.

1/2 = 3/6 and 1/3 = 2/6

So this  pair of fractions is 3/6, 2/6.

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9) It helps to draw a picture.  Since one angle is 60°, the other angle is 120°.  When you draw the ddiagonal lines, you will see that you end up with a 30°-60°-90° triangle where √3 is the hypotenuse. The longest diagonal will be across from the 60° angle, which will be length a√3. The length of the hypotenuse is 2a. (Remember: a, a√3, 2a are lengths for this special triangle)
2a = √3 → a = √3/2. 
side across from 60° = a√3 = (√3/2)·√3 → side from across 60° = 3/2
Remember that this is only one-half of the diagonal of the rhombus so multiply that by 2.
Length of longest diagonal is 3.

10) Volume of a square pyramid = (1/3) B · h, where B is the area of the base
200 = (1/3)B(12)
 50  = B
Since the base is a square, then each side is √50 (which equals 5√2).

Now, to find the lateral edge, use Pythagorean Theorem (a² + b² = c²).
Height (a) = 12,  base (b) = 5√2, and lateral edge (c) is unknown.

(12)² + (5√2)² = c²
144  +   50     = c²
     194           = c²
     √194         = c

Lateral edge is √194.




3 0
3 years ago
BRAINLIEST AND 50 PTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
frozen [14]

Answer:

The answer is A. Because 11/9 is greater than 1, the product of 4/5 and ​11/9 ​is greater than ​4/5

9/9 is exactly 1. 11/9 is greater than 9/9, so 11/9 is greater than 1. If I added 4/5, the answer is larger than 11/9, therefore the answer is A.

6 0
4 years ago
A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio, R, of the length of the s
Oksanka [162]

Solution :

Let the distance of the stick from one break be X

And let us assume that $X \leq l/2$.

Here, l = length of stick

Therefore, $P(X

We know that, $R=\frac{X}{l-X}$  ,  so by definition we get

$X =\frac{lR}{1+R}$

The cumulative distribution function for R is

$P(R

When it starts at zero, then r =0. It ends at one when the r has a maximum

value of one.

The probability density function is given by

\frac{d}{dr}(\frac{2r}{1+r})= \frac{2}{(r+1)^2}

Now integrating, we find E(R) and $E(R)^2$ gives :

$E(R) =\int\limits^1_0 \frac{2r}{(1+r)^2} \, dr = 2 \ln 2-1 $

$E(R)^2= 3 - 4\ln 2$

Therefore, Var(R)= $2-4(\ln \ 2)^2$

3 0
3 years ago
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