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kozerog [31]
3 years ago
15

The set of

Mathematics
1 answer:
joja [24]3 years ago
8 0

Answer:

B. Real numbers

Step-by-step explanation:

Real numbers are all numbers, rational and irrational, that could be found on a plane. This includes all positive and negative numbers, and all decimals. So, this only excludes imaginary numbers. The rest of the options exclude irrational numbers and more.

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What is the equation of the line that passes through (-9,12) and is perpendicular to the line whose equation is y=1/3x 6?
tigry1 [53]
Y = 1/3x ? 6....slope here is 1/3. A perpendicular line will have a negative reciprocal slope. All that means is " flip " the slope and change the sign. So our perpendicular line will have a slope of -3 (see how I flipped 1/3 and changed the sign)

y = mx + b
slope(m) = -3
(-9,12)...x = -9 and y = 12
now we sub and find b, the y int
12 = -3(-9) + b
12 = 27 + b
12 - 27 = b
-15 = b

so ur perpendicular equation is : y = -3x - 15
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3 years ago
The perimeter of a shape shows a car racing track, how far does a car travel in a race which consists of 23 laps 1.2km , 1.2 km
inna [77]

Answer:

110.4km

Step-by-step explanation:

1 lap is equal to 4.8

4.8×23=110. 4

(Need someone to confirm)

6 0
3 years ago
The entertainment committee voted to hire a DJ for the school dance. The DJ
nydimaria [60]
A 4 hours

This is because if you create the equation using $75 which is the starting fee and then $30 for every hour of playing music.

Equation: 75+30h<200
You subtract 75 from 200 and the divide that number by 200 you will get a decimal but you look at your answer choices and choose the closest number to your answer.
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3 years ago
Read 2 more answers
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
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Roman55 [17]

Answer:

Hello There. ☆~---_●₩●__---☆~ The students in each grade, that way it wont focus just on one group like the club, the teachers, or the boys.

Hope It Helps!~ ♡

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