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3 years ago

15y+3x-10=0 intercept form

Mathematics

1 answer:

storchak [24]3 years ago

5 0

Hai! So, I have all the math and explanations in the picture attached.
I forgot to add on there(and you'll see it) wen i subtract 3x and add 10 to the other side it cancels each other out. -3 + 3= 0 and -10 + 10 = 0. Etc.
Other than that, if you have any other questions, comment them down below. :)

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ser-zykov [4K]

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3 years ago

Whats the simplest form of 6 /18

MissTica

Answer:

What is 6/18 Simplified? - 1/3 is the simplified fraction for 6/18.

Step-by-step explanation:

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3 years ago

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I need help with this please. It’s probability.

Vinvika [58]

Answer:

2/13

Step-by-step explanation:

In a deck of 52 cards, there are 2 sets of red cards. Diamonds and Hearts. In each set of red cards, there are 4 fives. If there are 2 sets, that means that there are 8 red fives in the deck of 52 cards. This means that the probability is 8/52 which is simplified into 2/13.

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3 years ago

Evaluate the problem <br>

m_a_m_a [10]

Answer:

-2

Step-by-step explanation:

$Log_{2}\frac{1}{4}$ = -2

I hope this helps!

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2 years ago

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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

3 years ago