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Digiron [165]
2 years ago
11

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

Mathematics
1 answer:
olga2289 [7]2 years ago
4 0

Answer:

c.) 70

# carry on learning

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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

<u>- 0</u>: <u> 30</u>

12: -10

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

12:    -10

<u> - 1</u>: <u>+ 60 </u>

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 &gt; -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
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