Beaker A contains more water.
First,find equivalent fraction.
Next,compare the amounts you will see Beaker A contains more water.
Therefore,Beaker A contains more water.
Answer:
A. (-1, -16)
Step-by-step explanation:
Answer:
and
respectively.
Step-by-step explanation:
The given trigonometric equation is
We divide both sides by 
to get,
This implies that,
We take the inverse tangent of both sides to get,
Since the tangent ratio has a period of 
, another solution is
to the nearest degree.
In the third quadrant,
The solutions are
The value of x that satisfies the equation if x lies in the second quadrant is
The value of x that satisfies the equation if x lies in the third quadrant is
3(x - 2) < 18 and h + 6 > -3h + 12 would be graphed with an open circle.
Open circle is a term used for numbers that are less than (<) or greater than (>). On the other hand, Closed circle is used for numbers that are less than or equal to (≤ ) and greater than or equal to (≥).
Here, we are given four inequalities, let us check them one by one-
A. h + 6 > -3h + 12
⇒ 4h > 6
or h > 3/2
Here, we see that there is no equality sign, this means that 3/2 won't be included and hence h + 6 > -3h + 12 would be an open circle inequality.
B. 7 ≥ d/3
⇒ d ≤ 21
Here, we can clearly see the equality sign which means that 21 will be included. Hence, 7 ≥ d/3 would not be graphed with an open circle.
C. 3(x - 2) < 18
x - 2 < 6
x < 8
Here, we see that there is no equality sign, this means that 8 won't be included and hence 3(x - 2) < 18 would be graphed with an open circle.
D. 2y ≤ 14
y ≤ 7
Here, we can clearly see the equality sign which means that 7 will be included. Hence, 2y ≤ 14 would not be graphed with an open circle.
Thus, 3(x - 2) < 18 and h + 6 > -3h + 12 would be graphed with an open circle.
Learn more about inequalities here-
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I would like to see that “shaded region”
