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iVinArrow [24]
2 years ago
6

HELP!!! I can't figure out what do do or how to solve the last part!!

Mathematics
1 answer:
koban [17]2 years ago
3 0

Answer:

Use what you used for the first one and you should get the right answer.

Step-by-step explanation:

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The graph of f(x)=ax^2 is narrower than the graph of g(x)=dx^2 when a<d
Lynna [10]

Answer:

Step-by-step explanation:

as given in question that a > 0 so

if we put a=1

we get g(x) = f(x)

now put a =2

we get

g(x) = 2 f(x)

here we can see that g(x) would always be greater than or equals to f(x)

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Step-by-step explanation:

6 0
3 years ago
Prove the following statement using a proof by contraposition. Yr EQ,s ER, if s is irrational, then r + 1 is irrational.
sp2606 [1]

Answer:

I think that what you are trying to show is:  If s is irrational and r is rational, then r+s  is rational. If so, a proof can be as follows:

Step-by-step explanation:

Suppose that r+s is a rational number. Then r and r+s can be written as follows

r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0

r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0

Hence we have that

r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}

Then

s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}

This is a contradiction because we assumed that s is an irrational number.

Then r+s must be an irrational number.

3 0
2 years ago
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