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3 years ago

Trisha went to the store. She bought 3 kilograms of Swiss chard and 1 kilogram of Chinese

Mathematics

1 answer:

poizon [28]3 years ago

4 0

Answer:

1.99

Step-by-step explanation:

Swiss chard: $0.49 per kg

Chinese lettuce: $0.52 per kg =

= 3 kg Swiss chard + 1 kg Chinese lettuce

= 3 kg * $0.49/kg + 1 kg * $0.52/kg

= $1.47 + $0.52

= $1.99

Answer: 1.99

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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

Answer ,how to do it.

guapka [62]

Answer:

The system has two solutions:

x₁ = 5 ⇒ y₁ = -10

x₂ = -2 ⇒ y₂ = 11

Step-by-step explanation:

Let's solve the system of equations, this way:

y = -3x + 5

y = x ² - 6x - 5

Replacing y in the 2nd equation:

y = x ² - 6x - 5

-3x + 5 = x ² - 6x - 5

x ² - 3x - 10 = 0

Solving for x, using the quadratic formula:

(3 +/- √(9 -4 * 1 * -10))/2 * 1

(3 +/- √9 + 40)/2

(3 +/- √49)/2

(3 +/- 7)/2

x₁ = 10/2 = 5

x₂ = -4/2 = -2

x₁ = 5 ⇒ y₁ = -10

x₂ = -2 ⇒ y₂ = 11

As we can see the system has two different solutions

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sergiy2304 [10]

Answer:

B is the better detergent

Step-by-step explanation:

If A has a lower number on the 1-10 scale, that means on average it was closer to the lower end, or the least clean end. This means that A cleans worse than B., and B is better.

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leva [86]

Ana is 19 and Jordon is 31.

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How many whole servings of 3/4 cups are in 31/2 cups of ice cream?

IrinaVladis [17]

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