For
solving system of equations, we can use either substitution where we plug one
equation into the other, or elimination where we combine the equations.
-
Using elimination,
you would to eliminate one variable from both equations, so you automatically would
get one equation with one variable!
- Using
substitution
means you are going to solve one equation for one variable and substitute with
its value in the other equation in order to get also an equation with one
variable.
Let's take an example ...
y+x=2 and y-2x = 1
<span>Using <span>elimination, we need to subtract these two equation; one from the other...
y+x=2
-
y-2x=1
-----------
0+3x=1
then
x=1/3 and then substitute into any equation to get y-value</span></span>
y+x=2
y+1/3 = 2 >>>>>
y=5/3NOW...<span>Using substitution
</span>y+x=2 and y-2x = 1 >>(y=1+2x)
Plug (y=1+2x) into y+x=2 and solve for x
y+x=2
(1+2x) + x =2
1+3x = 2
3x=1
again (and for sure)
x = 1/3plug in x=1/3 into any of the equations above to get y:
y+x=2
y+1/3=2
y=5/3DOne !!!!!!
I hope you got
the idea
If you still need help, just let me know.
Answer:
Rewrite using the commutative property of multiplication.
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Step-by-step explanation:
The volume of the solid is 626.12m³
<h3>Volume of composite solids.</h3>
The volume of the given composite solid = volume of cylinder - volume of cube
The volume of cylinder Vc = πr²h
Vc = 3.14(8/2)²(13)
Vc = 3.14(16)(13)
Vc = 653.12 m³
Similarly the volume of the cube:
Volume of cube = 3³
Volume of cube = 27 m³
Volume of the solid = 626.12m³
Hence the volume of the solid is 626.12m³
Learn more on volume of composite solid here; brainly.com/question/9076728
Volume of the first aquarium = (12" x 13" x 14") = 2,184 inches³
Volume of the second aquarium = (32" x 34" x 35") = 38,080 inches³
The second aquarium has (38,080 / 2,184) = 17.44 times as much
volume as the first aquarium has.
So it'll take the hose 17.44 times as long to fill the second one
= (17.44) x (2 minutes) = 34.9 minutes.
Dimensions of the room in cm = 2.54 x 12 by 15 x 2.54 by 2.54 x 8.5 = 30.48 by 38.1 by 21.59
Volume of the room in cubic cm = 30.48 x 38.1 x 21.59 cubic cm = 25,072.21 cubic cm
Given that the density of air at room temperature is
, thus the mass of air in the room = 25,072.21 x 0.00118 = 29.59 g = 0.0296 kg
Given that the lethal dose of HCN is approximately 300 mg HCN per kilogram of air when inhaled, thus the <span>amount of HCN that gives the lethal dose in the small laboratory room is given by 300 x 0.0296 =
8.88 mg.</span>
