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2 years ago

Help me quick I need a good grade

Mathematics

1 answer:

Rasek [7]2 years ago

6 0

1.) This one is easy. The answer to this one is that the input is being multiplied by 3.

Proof: 1 x 3 = 3

2 x 3 = 6

3 x 3 = 9

2.) The answer is 12 because 4 x 3 = 12.

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This penny will double every day for the next 30 days. What is the answer? Can someone tell me at least how to set up the doubli

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2 years ago

Solve. Check your solution. -4+s= -68

Andrews [41]

-64 + -4 = -68

That is because 64 + 4 = 68 and the negatives in this case are the same way as if it was all positive.

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2 years ago

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My last problem and I'm stuck please help!!

WITCHER [35]

Answer is C= -125,125

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2 years ago

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by b :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since a/b < c/d, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by b/d to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (a + c)/(b + d) < c/d.

For the other side, you can do something similar and divide through everything by d :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and a/b < c/d tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (a + c)/(b + d) > a/b.

Then together we get the desired inequality.

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2 years ago

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Eduardwww [97]

Scatter plot II is the answer :)

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2 years ago

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