7/8
3/4
I'm order for them to have the same denominator you need to multiply so:
¾×2=6/8
So she drank 6/8 pints of water
The answer would be B and D. This can be found by putting the x and y values into the inequality to find whether or not they are true.
The given expression is

We have to find the discriminant first . And for that, first we need to move whole terms of the left side to right side, that is


The formula of discriminant is

Substituting the values of a,b and c, we will get

And since the discriminant is greater than 0, or it is positive so we have two real roots.
Therefore the correct option is B .
Check the picture below on the left-side.
we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.
now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.
so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.

![\bf \textit{area of a segment of a circle}\\\\ A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta ) \right] \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ ------\\ r=6\\ \theta =120 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20segment%20of%20a%20circle%7D%5C%5C%5C%5C%0AA_y%3D%5Ccfrac%7Br%5E2%7D%7B2%7D%5Cleft%5B%5Ccfrac%7B%5Cpi%20%5Ctheta%20%7D%7B180%7D~-~sin%28%5Ctheta%20%29%20%20%5Cright%5D%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0A%5Ctheta%20%3Dangle~in%5C%5C%0A%5Cqquad%20degrees%5C%5C%0A------%5C%5C%0Ar%3D6%5C%5C%0A%5Ctheta%20%3D120%0A%5Cend%7Bcases%7D)
Answer: y = -3(x +
)² +
,
,
<u>Step-by-step explanation:</u>
First, you need to complete the square:
y = -3x² - 5x + 1
<u> -1 </u> <u> -1 </u>
y - 1 = -3x² - 5x
y - 1 = -3(x² + 
y - 1 + -3(
) = -3(x² +
+
)
↑ ↓ ↑
= 
y - 1 -
= -3(x +
)²
y -
-
= -3(x +
)²
y -
= -3(x +
)²
y = -3(x +
)² +
Now, it is in the form of y = a(x - h)² + k <em>where (h, k) is the vertex</em>
Vertex =
,