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Bess [88]
3 years ago
6

In Exercise,find the horizontal asymptote of the graph of the function. f(x) = 2-xx

Mathematics
1 answer:
Dafna1 [17]3 years ago
6 0

Answer:

y=2

Step-by-step explanation:

Given the function y = 2-x ^ 2, we have that the horizontal asymptote is y = 2 since for x = 0 y = 2, that is, it is a line parallel to the axis = 0X where y has a value k, in this case y = 2

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Answer:

6 years

Step-by-step explanation:

t=2250/375

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Solve questions 1-10 I just need the answer in (x,y) form
Andre45 [30]

Answer:

1. (1,1)

2.(2,-6)

3.Not sure.

4.No solution

5.(-3,4)

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Step-by-step explanation:

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Step-by-step explanation:

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2 years ago
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PLEASE I need help on these 4 problems you don’t have to do them all but if you can at least do one of them TYSM
Brrunno [24]

Answer:

For right angle triangle,

we use Pythagoras theorem that is:

c^{2} =a^{2} +b^{2}

c = \sqrt{a^{2} +b^{2} }

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c = ?

a = 40

b = 9

putting them in formula,

c = \sqrt{40^{2} + 9^{2} }

c = 41

For question 2:

c = ?

a = 12

b = 13

putting them in formula,

c = \sqrt{12^{2} + 13^{2} }

c = approximately 17.69181

For question 3:

c = 35

a = 20

b = ?

putting them in formula,

c^{2} =a^{2} +b^{2}

35^{2} = 20^{2} + b^{2}

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b^{2} = 1225 - 400

b^{2} = 825

\sqrt{b^{2} } = \sqrt{825}

b = 5 \sqrt{33}

For question 4:

c = 37

a = 20

b = ?

putting them in formula,

c^{2} =a^{2} +b^{2}

37^{2} = 20^{2} + b^{2}

1369 = 400 + b^{2}

b^{2} = 1369 - 400

b^{2} = 969

Taking square root on both sides

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Hope it helps.

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2 years ago
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Step-by-step explanation:

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