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ikadub [295]
3 years ago
12

Nelson Mandela was a South African politician known for fighting racism and inequality. He was president of South Africa until h

e stepped down in the year 199919991999. Let xxx represent any year. Write an inequality in terms of xxx and 199919991999 that is true only for values of xxx that represent years after the year that Nelson Mandela stepped down from his presidency.
Mathematics
2 answers:
user100 [1]3 years ago
6 0
One possible inequality would be

1999-x<0.

If you subtract any year after he stepped down from 1999, the year he stepped down, the result will be a negative number.
Lady bird [3.3K]3 years ago
4 0
We know Nelson was president in the year 1999. If x, represents any year at all then we have to write an equality to represent years after the year that Nelson stepped down from the presidency.

1999-x<0

Because 1999 has to be less than the value it is up for. Any year after 1999 would be greater than 1999. This has to be a negative number because if you subtract any year after he stepped down the result will have to be negative, because the year will be after 1999. Hope this helps!
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Create an equivalent expression for 13 ^−7 x 13 ^5.
Sveta_85 [38]

13^{-7} \times 13^5=13^{-2}=\boxed{\frac{1}{13^2}}

8 0
1 year ago
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When x = 3 and y = 5, by how much does the value of 3x2 – 2y exceed the value of 2x2 – 3y ?
SpyIntel [72]

Answer:

14

Step-by-step explanation:

{3x}^{2}  - 2y \\ 3 { \times 3}^{2}  - 2 \times 5 \\ 27 - 10 = 17

{2x}^{2}  - 3y \\ 2 \times  {3}^{2}  - 3 \times 5 \\ 12 - 15 =  - 3

17 - 3 = 14

7 0
2 years ago
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Which point could be removed in order to make the relation a function? (–9, –8), (–{(8, 4), (0, –2), (4, 8), (0, 8), (1, 2)}
stepan [7]

We are given order pairs  (–9, –8), (–{(8, 4), (0, –2), (4, 8), (0, 8), (1, 2)}.

We need to remove in order to make the relation a function.

<em>Note: A relation is a function only if there is no any duplicate value of x coordinate for different values of y's of the given relation.</em>

In the given order pairs, we can see that (0, –2) and (0, 8) order pairs has same x-coordinate 0.

<h3>So, we need to remove any one (0, –2) or (0, 8) to make the relation a function.</h3>
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2 years ago
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Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec
julia-pushkina [17]

Answer:

(2,6,6) \not \in \text{Span}(u,v)

(-9,-2,5)\in \text{Span}(u,v)

Step-by-step explanation:

Let b=(b_1,b_2,b_3) \in \mathbb{R}^3. We have that b\in \text{Span}\{u,v\} if and only if we can find scalars \alpha,\beta \in \mathbb{R} such that \alpha u + \beta v = b. This can be translated to the following equations:

1. -\alpha + 3 \beta = b_1

2.2\alpha+4 \beta = b_2

3. 3 \alpha +2 \beta = b_3

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for \alpha,\beta and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

-\alpha + 3 \beta = 2

2\alpha+4 \beta = 6

whose unique solutions are \alpha =1 = \beta, but note that for this values, the third equation doesn't hold (3+2 = 5 \neq 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

-\alpha + 3 \beta = -9

2\alpha+4 \beta = -2

whose unique solutions are \alpha=3, \beta=-2. Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

4 0
2 years ago
jenette ordered a large bucket of popcorn at the movies.the graph shows the amount of popcorn in her bucket over time
Elenna [48]

Answer:

A

Step-by-step explanation:

5 0
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