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2 years ago

This is pretty easy, give me some synonyms for the word brazeness. When you do that find me some antonyms.

Mathematics

1 answer:

Alecsey [184]2 years ago

3 0

Answer:

Synonyms: Shameless, bold, audacious

Antonyms: Timid, shy

Step-by-step explanation:

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A student visiting the Sears Tower Skydeck is 1353 feet above the ground. Find the distance the student can see to the horizon.

kumpel [21]

601 miles ok cause it goes from the area by "A=BxH or A=BH

4 0

2 years ago

Read 2 more answers

Emily spent 1/2 of her money at the grocery store.Then, she spent 1/2 of what was left at the bakery. Next, at the music store,

Artist 52 [7]

I believe the answer is $48 because 6 * 2 is 12 and 12 * 2 is 24 and 24 * 2 is 48, thus 48$. Please feel free to check my work, though.

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3 years ago

Two lines have the given equations. Use the LINEAR COMBINATION METHOD to solve.

Maurinko [17]

Using linear combination method, the solution to given system of equations are (-7, -15)

<h3><u>Solution:</u></h3>

Linear combination is the process of adding two algebraic equations so that one of the variables is eliminated

Addition is used when the two equations have terms that are exact opposites, and subtraction is used when the two equations have terms that are the same.

<u><em>Given system of equations are:</em></u>

2x - y = 1 ---- eqn 1

3x - y = -6 ------ eqn 2

Subtract eqn 2 from eqn 1

2x - y = 1

3x - y = -6

(-) -------------

-x = 7

Substitute x = -7 in eqn 1

2(-7) - y = 1

-14 - y = 1

y = -14 - 1 = -15

Thus the solution to given system of equations are (-7, -15)

8 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

2 years ago

What is the volume, in cubic inches, of one cube with an edge length of 1/3 inch

DanielleElmas [232]

Answer:

1/27 cubic in.

Step-by-step explanation:

1/3*1/3*1/3= 1/27

3 0

2 years ago