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4 years ago

Philip ran 4 3/8 miles yesterday. Michael ran 1 5/8 miles yesterday.

Mathematics

1 answer:

Yakvenalex [24]4 years ago

5 0

Answer:

2 7/8

Step-by-step explanation:

You might be interested in

Find the difference between 1.97 and 20

cestrela7 [59]

Answer:

1.97-20= -18.03

20-1.97= 18.03

Step-by-step explanation:subtract the numbers from each other.

6 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

How many copies of 1 twelfth are equivalent to 3 sixths? Help!

andrezito [222]

Answer:

6/12 is equal to 3/6 ( 6 twelfths is equal to 3 sixths )

Step-by-step explanation:

multiply both sides of 3/6 by 2 to get the same denominator, doing that will get you 6/12

3 0

3 years ago

Read 2 more answers

If summer is in 277 days, 11 hours, 25 minutes, and 30 seconds. how long would it be in milliseconds?

Yakvenalex [24]

Answer:

24032430000

Step-by-step explanation:

in 277 days there are 23932800000 in 11 hours there are 39600000 and in 30 seconds there are 30000 so 23932800000+39600000+30000=24032430000

8 0

3 years ago

Marta runs a mandarin-orange fruit stand. She is talking to an employee, Maribel, on the phone and trying to tell her the equati

Ivanshal [37]

Answer:

Fifty more than half the number of the oranges is one-hundred-and-twenty.

The sum of fifty and half the number of oranges is the same as one-hundred-twenty.

Step-by-step explanation:

The equation is 50 + one-half m = 120.

Assume the number of oranges=m

50+1/2m=120

1/2m=120-50

1/2m=70

m=70÷1/2

=140

All statement that applies includes:

1. Fifty more than half the number of the oranges is one-hundred-and-twenty.

50+1/2m=120

2. The sum of fifty and half the number of oranges is the same as one-hundred-twenty.

50+1/2m=120

Martha can use the two statements above to communicate the correct equation

8 0

3 years ago

Read 2 more answers