In each box is a step in the process of solving a max/min problem like the one in Question 8. Number these steps in order (1-8)
1 answer:
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Answer:
a)TRUE
b)FALSE
c)TRUE
d)FALSE
e)TRUE
f)TRUE
g)TRUE
h)FALSE
i)FALSE
j)TRUE
Step-by-step explanation:
a) For every x there is y such that :
TRUE
This statement is true, because for every real number there is a square number of that number, and that square number is also a real number. For example, if we take 6.5, there is a square of that number and it equals 39.0625.
b) For every x there is y such that :
FALSE
For example, if x = -1, there is no such real number so that its square equals -1.
c) There is x for every y such that xy = 0
TRUE
If we put x = 0, then for every y it will be xy=0*y=0
d)There are x and y such that
FALSE
There are no such numbers. If we rewrite the equation we obtain an incorrect statement:
e)For every x, if there is y such that xy=1:
TRUE
The statement is true. If we have a number x, then multiplying x with 1/x (Since x is not equal to 0 we can do this for ever real number) gives 1 as a result.
f)There is x for every y such that if then xy=1.
TRUE
The statement is equivalent to the statement in e)
g)For every x there is y such that x+y = 1
TRUE
The statement says that for every real number x there is a real number y such that x+y = 1, i.e. y = 1-x
So, the statement says that for every real umber there is a real number that is equal to 1-that number
h) There are x and y such that
We have to solve this system of equations.
From the first equation it yields x=2-2y and inserting that into the second equation we have:
Which is obviously false statement, so there are no such x and y that satisfy the equations.
FALSE
i)For every x there is y such that
We have to solve this system of equations.
From the first equation it yields and inserting that into the second equation we obtain:
Inserting that back to the first equation we obtain
So, there is an unique solution to this equations:
x=1 and y=1
The statement is FALSE, because only for x=1 (and not for every x) exists y (y=1) such that
j)For every x and y there is a z such that
TRUE
The statament is true for all real numbers, we can always find such z. z is a number that is halway from x and from y.