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BigorU [14]
4 years ago
8

Write the sum of the numbers as the product of their gcf and another sum Problem: 30+54

Mathematics
1 answer:
omeli [17]4 years ago
4 0
The correct answer here is that the sum has to be 84. This is due to the fact that 30 + 54 is 84, and the greatest common factor is 6 because the factors of 30 are 1, 2, 3, 5, 6 , 10, 15, and 54's Factors: 1, 2, 3, 6, 9, 18, 27, and 54, so from this we can see that 6 is the great common factor.
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Point Q is the center of dilation. Line segment L M is dilated to created line segment L prime M prime. The length of L Q is 4 a
koban [17]

Answer:

The length of segment QM' = 6

Step-by-step explanation:

Given:

Q is the center of dilation

Pre-image (original image) = segment LM

New image = segment L'M'

The length of LQ = 4

The length of QM = 3

The length of LL' = 4

The original image was dilated with scale factor = 2

QM' = ?

To determine segment QM', first we would draw the diagram obtained from the given information.

Find attached the diagram

When a figure is dilated, we would have similar shape in thus cars similar triangles.

Segment L'M' = scale factor × length of LM

Let LM = x

L'M' = 2x

Using similar triangles theorem, ratio of their corresponding sides are equal.

QM/LM = QM'/L'M'

3/x = QM'/2x

6x = QM' × x

Q'M' = 6

The length of segment QM' = 6

5 0
3 years ago
Read 2 more answers
One package of socks cost $7. How many packages can you but with $56?
Akimi4 [234]

Step-by-step explanation:

$7 = 1 package socks

%56 = ??

$56/$7 × 1 package socks

= 8 × 1

= 8 package socks

Hope it helps!

4 0
3 years ago
Binomial Expansion/Pascal's triangle. Please help with all of number 5.
Mandarinka [93]
\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}

The rows add up to 1,2,4,8,16, respectively. (Notice they're all powers of 2)

The sum of the numbers in row n is 2^{n-1}.

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When n=1,

(1+x)^1=1+x=\dbinom10+\dbinom11x

so the base case holds. Assume the claim holds for n=k, so that

(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k

Use this to show that it holds for n=k+1.

(1+x)^{k+1}=(1+x)(1+x)^k
(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)
(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}

Notice that

\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}
\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}

So you can write the expansion for n=k+1 as

(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}

and since \dbinom{k+1}0=\dbinom{k+1}{k+1}=1, you have

(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}

and so the claim holds for n=k+1, thus proving the claim overall that

(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n

Setting x=1 gives

(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n

which agrees with the result obtained for part (c).
4 0
3 years ago
For the given set, first calculate the number of subsets for the set, then calculate the number of proper subsets. {18, 8, 14, 9
kicyunya [14]

Answer:

Subsets of the given set = 32

Proper subset of the given set = 31

Step-by-step explanation:

Given set is {18, 8, 14, 9, 6} having 5 elements.

We know number of subsets of a set having n elements are represented by

2^{n}

where n is the number of elements in the set.

Therefore, subsets of the given set = 2^{5}

= 32

Since original set itself is a subset of its own, is not a proper subset.

Therefore, proper subset of a set = 2^{n}-1

= 2^{5}-1

= 32 - 1

= 31

7 0
3 years ago
I need help on this.. And I will give you a BRANILIST if you the first one with the right answer ​
AlekseyPX

Answer:

0.1

Step-by-step explanation:

(9,4) (5,1)    \frac{y_{2} y_{1} }{x_{2} x_{1} }, plug in the numbers, \frac{(1)(4)}{(5)(9)}= \frac{4}{40} or \frac{1}{10}, to write it as a decimal, you divide the numerator by the denominator, 0.1

5 0
3 years ago
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