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4 years ago

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

1 answer:

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