HELP PLEASE !! Medal will be rewarded !

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

HELP PLEASE !! Medal will be rewarded !

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
plugin the numbers into the given equation ?
\(\large \sqrt{x-y} + m = d\)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

y is last two digits of ur birth year (u may fake if u want ;) ) m is month d is day
Okay
wat do u get for the equation ?
Let me work it out
okie
I can't get the answer .. I don't know how to work this out ! :(
if u can show me wat do u have... maybe i can help :)
did u plugin the values in equation ?
I don't know how to work it out .. Like I don't know where to start . I'm not good with radicals . All I did was try to find a way to work it out online but I couldn't find anything to help . Do I start with what's in the radical ?
nope, have u read the question ? it wants u to put ur date of birth in the given equation, and solve x
\(\large \sqrt{x-y} + m = d\) ^
Okay
put the year u born in that place
which year ?
Don't I use the values given in the question ?
Ohhh nevermind , I just read the question wrong . My bad
No, those values are for ur great great grandpa Nikola Tesla.... they're not urs lol
cool :)
Okay so I was born in 96
\(\large \sqrt{x-96} + m = d\) ^
there it goes ! wat about month and date
i feel like im interviewing u lol
Haha month is 4 and day is 22
thanks for giving it one shot :)
\(\large \sqrt{x-96} + 4 = 22\)
Alright so now what to do :o
solve for x
start by subtracting 4 both sides
Okay
\(\large \sqrt{x-96} + 4 = 22\) subtract 4 both sides \(\large \sqrt{x-96} = 18\)
18
Yes ! next, square both sides
You mean square 96 and 18 ?? :o
\(\large \sqrt{x-96} + 4 = 22\) subtract 4 both sides \(\large \sqrt{x-96} = 18\) square both sides \(\large (\sqrt{x-96})^2 = (18)^2\)
like this..
since u r not happy wid radical, we're killing it by squaring both sides !
\(\large \sqrt{x-96} + 4 = 22\) subtract 4 both sides \(\large \sqrt{x-96} = 18\) square both sides \(\large (\sqrt{x-96})^2 = (18)^2\) \(\large x-96 = 324\)
Oh okay !
wats our next step ?
Add 96 to both sides ?
thats it !
Okay so that would be 420 ? (;
Yes ! good job !!
Yay :D So that's the answer , correct ? Now what does it mean when it asks if the solution is extraneous ?? :o
oh, it asks to check if the solution is extraneous or now.
*not
Yeah , I don't know what that means .
its very easy to check, simply put the solution u got for x in the original (ur bday ) equaiton
Alright
ur bday equation :- \(\large \sqrt{x-96} + 4 = 22\) put x = 420 here, IF this satisfies this equation, then its NOT an extraneous solution.
I worked it out and I got 420=324 :o
careful, check again
But that's not true , so does that mean the solution IS extraneous ?
Oh okay ..
i want u check again,
Checking if x=420 is extraneous solution or not :- \(\large \sqrt{x-96} + 4 = 22\) put x = 420 \(\large \sqrt{420-96} + 4 = 22\)
Right
Checking if x=420 is extraneous solution or not :- \(\large \sqrt{x-96} + 4 = 22\) put x = 420 \(\large \sqrt{420-96} + 4 = 22\) \(\large \sqrt{324} + 4 = 22\)
Checking if x=420 is extraneous solution or not :- \(\large \sqrt{x-96} + 4 = 22\) put x = 420 \(\large \sqrt{420-96} + 4 = 22\) \(\large \sqrt{324} + 4 = 22\) \(\large 18 + 4 = 22\)
Ohhhh okay ! That makes more sense !
THANK YOU
Checking if x=420 is extraneous solution or not :- \(\large \sqrt{x-96} + 4 = 22\) put x = 420 \(\large \sqrt{420-96} + 4 = 22\) \(\large \sqrt{324} + 4 = 22\) \(\large 18 + 4 = 22\) \(\large 22 = 22\) TRUE, so x=420 is NOT an extraneous solution.
np :) yw !

Not the answer you are looking for?

Search for more explanations.

Ask your own question