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1 year ago

A rectangle has side lengths, L and W, and a diagonal, d.

Mathematics

2 answers:

Dafna11 [192]1 year ago

5 0

18.6 (Second questions)

According to the thyoreme of Pythagor, the side of a traingle squared plus the other side of the triangle squared is equal to the hypothenuse squared. The hypothenuse is the longer side of a triangle and in this case our diagonal. Then we find the square root of 15(2) + 11(2). It is 18.60349683.

That long number is equal to 18.6

<u><em>PLEASE MARK BRAINLIEST IF IT HELPED</em></u>

Rina8888 [55]1 year ago

3 0

Answer:

Step-by-step explanation:

w^2 + l^2 = d^2

if length = 15 and width =11

(11)^2 + (15)^2 = d^2

346 = d^2

$\sqrt{346}$ = d

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A group of 2 adults and 4 children spent $38 on tickets to a museum. A group of 3 adults and 3 children spent $40.50 on tickets

mrs_skeptik [129]

Answer:

adult $8.00

child $5.50

Step-by-step explanation:

Let the price of 1 adult ticket = x.

Let the price of 1 child ticket = y.

2x + 4y = 38

3x + 3y = 40.5

Multiply the first equation by 3. Multiply the second equation by -2. Then add them.

6x + 12y = 114

(+) -6x - 6y = -81

-----------------------------

6y = 33

y = 33/6

y = 5.5

2x + 4y = 38

2x + 4(5.5) = 38

2x + 22 = 38

2x = 16

x = 8

Answer:

adult $8.00

child $5.50

4 0

2 years ago

Read 2 more answers

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).

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Step-by-step explanation: