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AfilCa [17]
3 years ago
14

Will give Metal! From 1948 to 1994, South Africa existed under a system of apartheid. Under this system, white South Africans- a

bout 18% of the population- controlled all of the instruments of government: the Parliament, the Executive Branch, and the courts. Black and non-white South Africans had few legal rights and could not vote in elections. Voting was limited to white South African citizens. . This passage is describing what type of government? A) autocratic B) democratic C) dictatorship D) oligarchic,
Mathematics
1 answer:
TiliK225 [7]3 years ago
3 0
<span>Oligarchic An oligarchy is a small group of people having control of a country, organization, or institution. White South Africans were a very small percentage of the population, but had control over the entire government.</span>
You might be interested in
If y varies inversely as X and Y equals 5 when x equals 3 find X when Y is 15​
omeli [17]

Answer:

x = 1

Step-by-step explanation:

Given that y varies inversely as x then the equation relating them is

y = \frac{k}{x} ← k is the constant of variation

To find k use the condition y = 5 when x = 3

k = yx = 5 × 3 = 15, thus

y = \frac{15}{x} ← equation of variation

When y = 15 then

15 = \frac{15}{x} ( multiply both sides by x )

15x = 15 ( divide both sides by 15 )

x = 1

5 0
3 years ago
An artist wishes to make a painting in the shape of a golden rectangle, based on the theory that this shape is the most pleasing
Arisa [49]
C because 9 x 12 is 108 and so it’s c but I’m sorry if this is wrong
5 0
3 years ago
A line that passes through the point (4, 2) and has the slope of -3/2
mash [69]
The answer is: y = -3/2 x + 8

Graph:

To find the answer we use the formula;

y-y1 = M1 (x - x1)

y-2 = -3/2 (x-4)

When fully solved: the answer is:
y = -3/2 x + 8

8 0
2 years ago
Which is NOT TRUE of data sets? A) A measurable characteristic of a sample is called a statistic. B) A measurable characteristic
zheka24 [161]
A) is wrong because a measurable characteristics is called a basic

3 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
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