All categories

3 years ago

2) 0.275 is what type of number? Check all that apply. Rational Integer Natural Real

Mathematics

2 answers:

Vadim26 [7]3 years ago

8 0

Answer:

Rational

Real

Step-by-step explanation:

A rational number in which the decimal doesn't repeat forever.

A real number is any number that isn't imaginary.

The number 0.275 checks those two categories.

An integer is a number that is not a fraction.

A natural number is any positive number that is not a decimal.

The number 0.275 does not check these two categories.

Best of Luck!

iogann1982 [59]3 years ago

6 0

Answer:

A) Rational

Step-by-step explanation:

0.275 is a rational number.

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Multiply 3/5 and 3/4

ella [17]

Answer:

6/9

Step-by-step explanation:

5 0

3 years ago

If A = {w.a, t,c, h, d, 0,8} and U = {a,b,c,d, e, f, g, h, L.), k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z), find A'

Ugo [173]

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3 years ago

A university is building a new student center that is two- thirds the distance from the arts center to the residential complex.

Natasha2012 [34]

Answer:

$C = (\frac{21}{5},\frac{33}{5})$

Step-by-step explanation:

Given

Points: (1, 9) and (9, 3)

Ratio = 2/3

Required

Determine the coordinate of the center

Represent the ratio as ratio

$Ratio = 2:3$

The new coordinate can be calculated using

$C = (\frac{mx_2 + nx_1}{n + m},\frac{my_2 + ny_1}{n + m})$

Where

$(x_1,y_1) = (1, 9)$

$(x_2, y_2) = (9, 3)$

$m:n = 2:3$

Substitute these values in the equation above

$C = (\frac{2 * 9 + 3 * 1}{3 + 2},\frac{2 * 3 + 3 * 9}{2 + 3})$

$C = (\frac{18 + 3}{5},\frac{6 + 27}{5})$

$C = (\frac{21}{5},\frac{33}{5})$

Hence;

<em>The coordinates of the new center is </em> $C = (\frac{21}{5},\frac{33}{5})$ <em></em>

7 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

3) The sum of three consecutive numbers is 72. What are the smallest of these numbers?

34kurt

Answer:

Step-by-step explanation:

23 + 24 + 25 = 72

6 0

3 years ago