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2 years ago

Compare these two functions, where the input variable represents days.

Mathematics

1 answer:

olga2289 [7]2 years ago

4 0

Answer:

c.) 70

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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

2 years ago

PJ will begin her cake deliveries at 12:20 . She asked to remind her 30 minutes before she has to leave. What time should I remi

mojhsa [17]

Answer:

At 11:50

Step-by-step explanation:

1. Subtract the hours and the minutes separately

12: 20

12: -10

2. If the minutes are negative, subtract 1 from the hours and add 60 to the minutes.

12: -10

11: 50

You should remind PJ about her deliveries at 11:50.

4 0

3 years ago

Can someone help me with this 4 - 3m > -5

ella [17]

Answer: M < -3

.................................................

Step-by-step explanation:

4 - 3m > -5

4 - 3m > (-4 -5)

3m > -9

(3/3)m > (-9/3)

m<-3

4 0

3 years ago

The cost of a hotel room during Nigel's trip is $180. The hotel room tax in the city he is in is 11.5%. What is the total cost o

kumpel [21]

Answer:

Step by Step:

multiply the cost by the percentage.

Then add the cost to the answer

Answer:200.70

6 0

3 years ago

Write the equation of a circle with a center at (1,-9) and a radius of 12

Andre45 [30]

The equation of the circle is (x - 1)^2 +(y + 9)^2 = 144

<h3>How to determine the circle equation?</h3>

The given parameters are:

Center, (a,b) = (1, -9)

Radius, r = 12

The circle equation is calculated as:

(x - a)^2 + (y- b)^2 = r^2

This gives

(x - 1)^2 +(y + 9)^2 = 12^2

Evaluate the exponent

(x - 1)^2 +(y + 9)^2 = 144

Hence, the equation of the circle is (x - 1)^2 +(y + 9)^2 = 144

Read more about circle equations at:

brainly.com/question/10618691

#SPJ1

8 0

1 year ago