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2 years ago

PLEASE PLEASE PLEASE HELPP!!??

Mathematics

2 answers:

Furkat [3]2 years ago

4 0

By the CORRESPONDING ANGLES POSTULATE:

if lines l and m are PARALLEL, then <1 ~= <5

converse:
if <1 ~= <5, then lines l and m are PARALLEL,

kolezko [41]2 years ago

3 0

Angle 1 equals angle 5. I think the theorem is Alternate interiors?

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Which type of graph would best show the population of New York City over 70 years

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Answer:

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2 years ago

Subtract five from three times a number and the result is equal to the number plus five. find the number.

allsm [11]

3(x)-5=x+5
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3x=x
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because 3x=x is basically would make each part into 0 which would end up the same.

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3 years ago

Factor<br> 2x2 - 13x - 70

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2 years ago

Read 2 more answers

Closed circle on 2 to closed circle on 6

Stella [2.4K]

Answer:

see the explanation

Step-by-step explanation:

we have

closed circle on 2 to closed circle on 6

That means -----> All real numbers greater than or equal to 2 and less than or equal to 6

In interval notation is ------> [2,6]

In set builder notation is ----> { x ∈ R | x ≥ 2 and x ≤ 6 }

7 0

2 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

4 0

2 years ago