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Grace [21]
3 years ago
9

write the sum of the numbers as the product of their greatest common factor and another sum for 15 81

Mathematics
1 answer:
velikii [3]3 years ago
8 0
15 + 81 = 3(5 + 27).
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lawyer [7]
I believe 3b/6a which could reduce to b/3a.
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What causes changes of states in matter?
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The answer is A I believe
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An ANOVA procedure is used for data obtained from four populations. Four samples, each comprised of 30 observations, were taken
tatiyna

Answer:

D. 3 and 116

Step-by-step explanation:

d.f.N = k - 1 (numerator degrees of freedom) = 4 - 1 = 3

N = 4 × 30 = 120

d.f.D = N - k (denominator degrees of freedom) = 120 - 4 =116

7 0
3 years ago
Find the complete perimeter of the sector that intercepts arc AC.
zhuklara [117]
To answer, determine first the length of the arc intercepted by the angle which measures 45°. 

Calculate for the circumference of the circle with the equation,
    C = 2πr
where C is circumference and r is radius. Substitute the known value for r,
    C = 2π(7) = 43.98

Then, multiply the obtained value by the ratio of the angle intercepted and the whole revolution. 
    Arc = 43.98 x (45°/360°)
    Arc = 5.5 

The whole perimeter of the sector includes the arc and 2 radii. Hence, the equation for the whole perimeter is,
   Perimeter = Arc + 2r
Substituting the values,
   Perimeter = (5.5) + 2(7)
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<em>ANSWER: 19.50 units</em>
7 0
3 years ago
Use a direct proof to show that the product of two odd integers is odd.
yulyashka [42]

Answer:

The proof itself

Step-by-step explanation:

We can define the set of all even numbers as

E = \{ a \in \mathbb{Z} \setminus a = 2.k , k\in \mathbb{N}\}

This is, we can define all even numbers as the set of all the multiples of 2

As for the odd numbers, we can always take every even number and sum one to each one. This is

O = \{ a\in \mathbb{Z} \setminus a=2.k+1,k\in\mathbb{N}_{0}\}

Note that k\in\mathbb{N}_{0}(the set of all natural numbers adding the zero) so that for k=0 then a=1

Now, given 2 odd numbers a and b we can write each one as follows:

a = 2k+1\\b = 2l+1\\k,l \in\mathbb{N}_{0}

And then if we multiply them with each other we obtain:

a.b = (2k+1).(2l+1)\\= 4kl+2k+2l+1\\= 2(2kl+k+l) + 1\\= 2k'+1 \\where\ k'=2kl+k+l

Then we have that a.b is also an odd number as we defined them.

6 0
3 years ago
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