Capacity=L^3
=8×8×8
=512L
Answer:
The set of all real numbers
Step-by-step explanation:
The domain is the set of all real numbers (both positives and negatives including zero.
OK. These problems are easy if you know the quadratic formula,
and they're impossible if you don't.
Here's the quadratic formula:
When the equation is in the form of Ax² + Bx + C = 0
then x = [ -B plus or minus √(B²-4AC) ] / 2A
I'm sure that formula is in your text or your study notes,
right before these questions. You should cut it out or
copy it, and tape it inside the cover of your notebook.
Then, you'll always have it when you need it, until
you have it memorized and can rattle it off.
The first question says 3x² + 5x + 2 = 0
Is this in the form of Ax² + Bx + C = 0 ?
Yes ! A=3 B=5 C=2
so you can use the quadratic formula to solve it.
x = [ -B plus or minus √(B²-4AC) ] / 2A
= [ -5 plus or minus √(5² - 4·3·2) ] / 2·3
= [ -5 plus or minus √(25 - 24) ] / 6
= [ -5 plus or minus √1 ] / 6
x = -4 / 6 = -2/3
and
x = -6 / 6 = -1 .
_______________________________________
The second question says
4x² + 5x - 1 = 0
Is this in the form of Ax² + Bx + C = 0 ?
Yes it is ! A=4 B=5 C= -1
so you can use the quadratic formula to solve it.
x = [ -B plus or minus √(B²-4AC) ] / 2A
Now, you take it from here.
The slope of the line is 2/3. the answer is C
Answer:
The remaining liquid has a volume of 345.4 cm³.
Step-by-step explanation:
To solve this question we first need to calculate the volume of each container.
Cylindrical:
Volume = h*pi*r² = 14*3.14*(5)² = 1099 cm³
Cone:
Volume = (1/3)*h*pi*r² = (1/3)*20*3.14*(6)² = 753.6 cm³
Since the liquid from the cylinder was transfered to the cone until the latter was full, then what remains in the cylinder is it's original volume minus the cone volume. So we have:
remaining liquid = 1099 - 753.6 = 345.4 cm³
