Rules for adding negative and positive signs
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ANSWER
Find out the how many 3, 4, and 5 point questions could there be.
To proof
Let us assume that the 3 points based question be = x
Let us assume that the 4 points based question be = y
Let us assume that the 5 points based question be = z
As given
An instructor wants to write a test with 25 questions
than the equation become in the form
x + y + z = 25
he quiz to be worth a total of 100 points.
than the equation is becomes
3x + 4y + 5z =100
As given
He wants the number of 3-point questions to be 4 less than the number of 4-point questions
x = y -4
Than the three equation are
x + y + z = 25 ,3x + 4y + 5z =100 and x = y -4
put x = y -4 in the x + y + z = 25 ,3x + 4y + 5z =100
than
2y + z = 29, 7y +5z= 112
multiply 2y + z = 29 by 5 and subtracted 7y +5z= 112
10 y -7y + 5y -5y = 145 -112
3y = 33
y =11
put in the x = y -4
x = 11-4
x= 7
put the value of x ,y in the x + y + z = 25
7 + 11 +z =25
18 +z = 25
z = 25 -18
z=7
therefore
numbers of the 3 point question be = 7
numbers of the 4 point question be = 11
numbers of the 5 point question be = 7
Hence proved
It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if
</span><span>
</span><span>In notation we write respectively
</span>
Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²<span>
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c</span>²
That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers