anonymous
  • anonymous
Solve for d.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
across
  • across
EVERYONE STOP NOW
across
  • across
qudrex, try solving this one here, and we will guide you through the process.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
okay
across
  • across
So, what is the first step?
anonymous
  • anonymous
no idea. :/ i fail hard at math.
across
  • across
You are trying to solve for \(d\) in\[\sqrt{d^2-11}=5.\]It would be a good idea to start by squaring both sides, do you agree?
anonymous
  • anonymous
that would give me -22 and 10 right?
across
  • across
Howcome? If you square both sides, you will get\[d^2-11=25.\]Do you see why?
anonymous
  • anonymous
no. :/
across
  • across
Let us go back one step: If we square\[\sqrt{d^2-11}=5,\]then we are doing this:\[\left(\sqrt{d^2-11}\right)^2=5^2,\]right?
anonymous
  • anonymous
yeah.
across
  • across
Do you know what\[\left(\sqrt{d^2-11}\right)^2\]is?
anonymous
  • anonymous
no.
across
  • across
The square root sign and the square cancel each other out to give you\[d^2-11.\]Do you know what \(5^2\) is?
anonymous
  • anonymous
10 right?
across
  • across
By definition, \(5^2=5\cdot5=?\)
anonymous
  • anonymous
so 25.
across
  • across
That is why you get\[d^2-11=25.\]Can you solve that for \(d\)?
anonymous
  • anonymous
d = 36? :/
across
  • across
You are close!\[d^2=36.\]We are not done yet, however: we need \(d\) not \(d^2\). Do you know how to fix that?
anonymous
  • anonymous
divide 36 by 2?
across
  • across
No. Just like before, the way to cancel out a square is by taking the square root. That is,\[\sqrt{d^2}=d.\]What you want to do here is\[\sqrt{d^2}=\sqrt{36}.\]What is that?
anonymous
  • anonymous
these are the multipule choice. 4, –4 5, –5 5, 6 6, –6
across
  • across
Oh, well...\[\sqrt{36}=\pm6.\]
anonymous
  • anonymous
so its 6, -6?
across
  • across
Yes.
anonymous
  • anonymous
okay, thankk yoou.
across
  • across
I hope you learned something, though.

Looking for something else?

Not the answer you are looking for? Search for more explanations.