The following are the answers to
the questions presented:
Part 1:
Vertex: <span>(<span><span>−3</span>/ 2</span>,<span> 11/2</span>)</span>
Axis of
symmetry = x = -3/2
Domain = all
real numbers
Range = <span>y </span><span>≤ </span>11/2
Part 2:
A quadratic equation is symmetrical around its vertex. The
equation for the axis of symmetry is x = -3/2 since the equation is in terms of
x. Since no value of x is undefined, then the domain of the equation is clearly
all real numbers. Since the value of “a” is negative, then that means the y
coordinate of the vertex is the maximum value so the range will never get above
11/2. I am hoping that these answers have satisfied your queries and it
will be able to help you in your endeavors, and if you would like, feel free to
ask another question.
Can you be more specific pls
D I got the phone call back in the phone with me a couple
The answer to this question would be: p+q+r = 2 + 17 + 39= 58
In this question, p q r is a prime number. Most of the prime number is an odd number. If p q r all odd number, it wouldn't be possible to get 73 since
odd x odd + odd= odd + odd = even
Since 73 is an odd number, it is clear that one of the p q r needs to be an even number.
There is only one odd prime number which is 2. If you put 2 in the r the result would be:
pq+2= 73
pq= 71
There will be no solution for pq since 71 is prime number. That mean 2 must be either p or q. Let say that 2 is p, then the equation would be: 2q + r= 73
The least possible value of p+q+r would be achieved by founding the highest q since its coefficient is 2 times r. Maximum q would be 73/2= 36.5 so you can try backward from that. Since q= 31, q=29, q=23 and q=19 wouldn't result in a prime number r, the least result would be q=17
r= 73-2q
r= 73- 2(17)
r= 73-34=39
p+q+r = 2 + 17 + 39= 58
I agree. The answer is choice B
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f ' (1) is shown to be positive
f ' (2) is shown to be positive as well
There are no critical values between x = 1 and x = 2, so f ' (x) is positive on the interval 1 < x < 2, so f(x) is increasing on this interval
All of this points to either choice A or choice B
--------------------------------
Similarly,
f ' (3) is negative
f ' (4) is negative
so f ' (x) is negative where 3 < x < 4 due to no other critical values being between x = 3 and x = 4
Based on these facts alone, the answer is either choice B or choice D
But we know that it can't be choice D as we determined it was between A and B. This rules out choice D
So all that's left is choice B
