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2 years ago

Good luck to anyone that has a week left of school and just now starting e2020 projects!

Mathematics

2 answers:

Veseljchak [2.6K]2 years ago

6 0

Answer:

thanks i need all the luck i can get seeing how i was walking my snail earlier and it got crushed by my shoe it was sad and thwn my turkey was eating it and then my turkey died cuz my snail was sick now my turtle is stuck on my roof hows that for a 2020 project

Zarrin [17]2 years ago

6 0

Answer:

to you too

Step-by-step explanation:

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Which domain restrictions apply to the rational expression?

Anton [14]

It can either be the negative sign of the number by the x so it should either be 2 or -

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2 years ago

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Fofino [41]

Tree 3 diagram is the answer

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2 years ago

(m? - 5 m +6 ) ÷(m - 2)

leonid [27]

Answer:

(m? - 5 m +6 ) ÷ (m - 2) = $\frac{m}{-5m+6} / (m - 2)$

Step-by-step explanation:

Divide the first expression by the second expression.

$\frac{m}{-5m+6} / (m - 2)$

I hope this helps.

4 0

2 years ago

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How many different ways are there to choose a subset of the set {1,2,3,4,5,6} so that the product of the members of the subset i

zvonat [6]

You have to pick at least one even factor from the set to make an even product.

There are 3 even numbers to choose from, and we can pick up to 3 additional odd numbers.

For example, if we pick out 1 even number and 2 odd numbers, this can be done in

$\dbinom 31 \dbinom 32 = 3\cdot3 = 9$

ways. If we pick out 3 even numbers and 0 odd numbers, this can be done in

$\dbinom 33 \dbinom 30 = 1\cdot1 = 1$

way.

The total count is then the sum of all possible selections with at least 1 even number and between 0 and 3 odd numbers.

$\displaystyle \sum_{e=1}^3 \binom 3e \sum_{o=0}^3 \binom 3o = 2^3 \sum_{e=1}^3 \binom3e = 8 \left(\sum_{e=0}^3 \binom3e - \binom30\right) = 8(2^3 - 1) = \boxed{56}$

where we use the binomial identity

$\displaystyle \sum_{k=0}^n \binom nk = \sum_{k=0}^n \binom nk 1^{n-k} 1^k = (1+1)^n = 2^n$

3 0

11 months ago

The midpoints of the hexagon are connected to form another hexagon, this pattern continues indefinitely. If the area of the 99th

jarptica [38.1K]

Answer:

(4/3)⁹⁸ units²

Step-by-step explanation:

Regular hexagons have a property that cutting off the triangles obtained by joining the midpoints of consecutive sides leaves a hexagon of 3/4 of the area.

This means the area of each hexagon inside out starting from the one with unit area is 4/3 of the previous one.

99 ⇒ 1
98 ⇒ 1*4/3
97 ⇒ 1*(4/3)²
...
1 ⇒ (4/3)⁹⁹⁻¹

The original hexagon has the area of:

(4/3)⁹⁸ units²

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2 years ago

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