I have no idea. Just commenting to get the answer when someone else knows it
Answer:
(9,5.5)
Step-by-step explanation:
= 2x^2 - 20x + 49 = 2(x^2 - 10x + 49/2)
y= 2((x-5)^2 - 25 + 49/2) = 2((x-5)^2 - 1/2) = 2(x-5)^2 - 1
q =5, r=-1, p=2
Vertex = (q,r) = (5,-1)
hope it helps plz mark as brainliest
10000 digits can be used for 4 digit A.T.M code.
<u>Solution:</u>
Given that A.T.M required 4 digit codes using the digits 0 to 9.
Need to determine how many four digit code can be used.
We are assuming that number starting with 0 are also valid ATM codes that means 0789 , 0089 , 0006 and 0000 are also valid A.T.M codes.
Now we have four places to be filled by 0 to 9 that is 10 numbers
Also need to keep in mind that repetition is allowed in this case means if 9 is selected at thousands place than also it is available for hundreds, ones or tens place .
First digit can be selected in 10 ways that is from 0 to 9.
After selecting first digit, second digit can be selected in 10 ways that is 0 to 9 and same holds true for third and fourth digit.
So number of ways in which four digit number is created = 10 x 10 x 10 x 10 = 10000 ways
Hence 10000 digits can be used for 4 digit A.T.M code.
16x - 5y = -33 First, you subtract 16x from both sides of the
<u>-16x -16x </u> equation
<u>-5y</u> = <u>-16x -33</u> Next, you divide both sides of the equation by -5 to
-5 -5 make "y =" form
y = 16/5x + 33/5
