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2 years ago

Monroe started saving money. Right now he

Mathematics

2 answers:

Ray Of Light [21]2 years ago

6 0

Answer:

After 4 months

Step-by-step explanation:

200+15+15+15+15=260

300-10-10-10-10=260

Alexeev081 [22]2 years ago

5 0

Answer:

They will have the same amount of money in exactly 4 months

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What’s the volume of a cylinder with a radius of 1 and height of 3ft?

Dafna1 [17]

Answer:

9.42

Step-by-step explanation:

V=pir2h

(pi, radius squared, height)

hope this helped!

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2 years ago

Help me please help pleaseeeeeeeeeeeee

LenaWriter [7]

Answer:

(8,3)

Step-by-step explanation:

Point c is seven units away from the line x = 1

So we move seven units to the right of the line x = 1

the x value = 8

y value = 3

Coordinates = (8,3)

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2 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

Select from the drop-down menus to correctly complete each statement. −20>−21 −19>−20 −21 is located to the of −20 on a nu

LekaFEV [45]

Answer:

left, right

Step-by-step explanation:

If you use the < or ≤ symbol to compare the numbers, their left-right order is the same as on the number line. Here, we would write the ordering as ...

-21 < -20 < -19

So, we see that ...

-21 is <em>left</em> of -20 on the number line

-19 is <em>right</em> of -20 on the number line

4 0

3 years ago

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Tell whether the ordered pair is a solution to the system.

DerKrebs [107]

Answer:

Step-by-step explanation:

no. substitute -1 for x and -2 for y.

the first one is -1 - 10

this equals -11 not 3

7 0

2 years ago

Read 2 more answers