anonymous
  • anonymous
BUM BUM BUMMMMMMM I NEED HELPPPPPP!!!! The figure shows the graph of the equation y =k–x2, where k is a constant. If the area of triangle ABC is 1/8, what is the value of k?
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1438987265534:dw|
anonymous
  • anonymous
can i help
wintersuntime
  • wintersuntime
I got 1/4, by taking the area of triangle 1/2xbasexheight=1/8. basexheight will be 1/4. Hence, k = 1/4

Looking for something else?

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

More answers

wintersuntime
  • wintersuntime
Do you need any more help ?
Loser66
  • Loser66
@wintersuntime base x height = 1/4, how can you know k =1/4? I agree that k =1/4 but don't understand your logic.
wintersuntime
  • wintersuntime
I'll be right back it's cuz im helping someone else
dumbcow
  • dumbcow
|dw:1438997626829:dw| \[A = \frac{1}{2}(2 \sqrt{k})(k) = \frac{1}{8}\] \[k^{3/2} = \frac{1}{8}\] \[k = \frac{1}{8^{2/3}} = \frac{1}{(\sqrt[3]{8})^2} = \frac{1}{4}\]
triciaal
  • triciaal
when x = 0 y = k

Looking for something else?

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