All categories

2 years ago

Is the statement “ for a set of whole numbers, the longest numeral will belong to the largest number” true or false? Why?

Mathematics

1 answer:

IRISSAK [1]2 years ago

5 0

Numerals are symbols and figures that are used to denote or represent numbers.

It is false that “for a set of whole numbers, the longest numeral will belong to the largest number”

<h3>How to determine the true statement</h3>

The length of a numeral does not determine the value of the number,

The above statement means that

The whole number with the longest numeral in a set may or may not be the largest.

Take for instance the numbers 8 and 9.

In Roman numerals,

8 is represented with VIII and 9 is represented with IX

The numeral VIII is longer than the numeral IX, but IX is greater than VIII

Hence, the statement is false

You might be interested in

HELP ME PLZZ I NEED HELP WITH THIS!!

VARVARA [1.3K]

Answer:

Step-by-step explanation:

7 0

3 years ago

Anthony travels from Page Az to Flagstaff Az at a rate of 60 miles per hour in 2.5 hours. How far is Flagstaff from Page?( Show

DerKrebs [107]

Answer:

150 miles

Step-by-step explanation: The distance from Page to Flagstaff can be shown with the equation d = r * t, or distance = rate * time. In this case, the rate is 60 mph, and the rate is 2.5 h. Plugging that into the equation you get d = 2.5(60) = 150, giving you your distance.

4 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

Hellllppppp meeeeeee plzzzzz fasttttt

valentinak56 [21]

Answer:

A rate: a measure, quantity, or frequency, typically one measured against some other quantity or measure.

A unit rate: When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates. If you have a multiple-unit rate such as 120 students for every 3 buses, and want to find the single-unit rate, write a ratio equal to the multiple-unit rate with 1 as the second term.

Hope this helps.

4 0

4 years ago

Write this number in word form

expeople1 [14]

Ok its it is three hundred million seven hundred sixty three one hundred thirty six

3 0

3 years ago

Read 2 more answers