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3 years ago

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The owner of a bike shop sells unicycles and bicycles and keeps inventory by counting seats and wheels. One day, she counts 21 s

eats and 30 wheels. Which equation in slope-intercept form shows that x bicycles and y unicycles together have a total of 30 wheels?
A. y = -2x - 30
B. y = -2x + 30
C. y = 2x + 30
D. y = 2x - 30

1 answer:

Alex787 [66]3 years ago

3 0

It would be b I hope I helped you

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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).

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3 years ago

Mary’s grade on her research paper counts for 20% of her final grade in english. if there are a total of 400 points possible how

sweet-ann [11.9K]

Answer:

80 points

Step-by-step explanation:

400 x 0.20% = 80

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3 years ago

Ron walked 8/10 mile from his grandmothers house to store and then he walked 9/10 mile to his house use benchmarks to estimate a

Neporo4naja [7]

8/10+9/10=17/10=1 7/10=2
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Suppose a city with population 800,000 has been growing at a rate of 5% per year. If this rate continues, find the population of

AleksandrR [38]

Answer:

2,340,208.5 or 2,340,209

Step-by-step explanation:

You could answer this by multiplying the current population of 800,000 by 1.05 (1.05 represents the annual growth rate of population) and do that 22 times. But that would take a while.

we got 1.05 because the formula says that y = a( 1 + r ) power t

for a we will give it 1

so y = 1 ( 1 + 0.05) power t

it will give us 1.05

to do this faster, you would first calculate (1.05)22and then multiply this by 800,000.

So, 800,000 x (1.05)22 = 2,340,208.5 or 2,340,209

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2 years ago

What is the area of this circle in square inches? Use 3.14 for n. Round your answer to the nearest tenth.

Lostsunrise [7]

Answer: 615.4

Step-by-step explanation:

$A=\pi r^{2}$

if the radius is 14

$A=\pi 14^{2} \\A=196\pi \\A=196(3.14)\\A=615.44\\$

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3 years ago

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