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Lostsunrise [7]
2 years ago
11

PLSSS HELPPP I WILL GIVE BRAINLIESTTTT!!!!!!!

Mathematics
1 answer:
Ksenya-84 [330]2 years ago
8 0

Answer:

yes to both

Step-by-step explanation:

For a relationship to be a function then each value of x must relate to exactly one unique value of y.

This is the case in this table as

2 → 76

4 → 152

6 → 228

All unique values thus a function

To determine if it is linear then the rate of change is constant for each set of ordered pairs. This is measured using the slope formula.

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (2, 76) and (x₂, y₂ ) = (4, 152 ) ← ordered pair from table

m = \frac{152-76}{4-2} = \frac{76}{2} = 38

Repeat with (x₁, y₁ ) = (4, 152) and (x₂, y₂ ) = (6, 228)

m = \frac{228-152}{6-4} = \frac{76}{2} = 38

Since rate of change is constant then function is linear

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If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn
mash [69]

Let a=\tan A and b=\sin A. Then

m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab

\implies m^2-n^2=4\tan A\sin A

and

mn=(a+b)(a-b)=a^2-b^2

\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}

The expression under the square root can be rewritten as

\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)

Recall that

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so that

\tan^2A-\sin^2A=\sin^2A\tan^2A

and assuming \sin A>0 and \tan A>0, we end up with

4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A

so that

m^2-n^2=4\sqrt{mn}

as required.

5 0
3 years ago
Write in least to greatest order 0.56, 4.56, 0.65
SIZIF [17.4K]
Hey! The answer is: 0.56, 0.65, 4.56 ! Have a great day, bud!
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Is that the answer please help no links no links please help
fenix001 [56]

Answer:

360

Step-by-step explanation:

V= L•W•H/3

10×12×9÷3

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2 years ago
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dangina [55]
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Hope this helped

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7 0
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