Answer:
The solution of a linear inequality is the ordered pair that is a solution to all inequalities in the system and the graph of the linear inequality is the graph of all solutions of the system. Graph one line at the time in the same coordinate plane and shade the half-plane that satisfies the inequality
Answer:
Step-by-step explanation:
Local minimum. It would be the minimum value of the function not including the end points. If you are including the end points, its an absolute minimum
I think you mean y=ab^x.
First, plug in the values of x=2 and y=24 for the first equation, and x=3 and y=48 for the second one:
24 = ab^2
48 = ab^3
divide equation 2 by equation 1:
2 = b^1
or, more simply:
2 = b
now plug b into one of the equations to get "a":
24 = a*2^2
24 = a*4
6 = a
So, the answer is y = 6(2)^x
(I think you need to specify the power of x in your answers)
-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are blue-green-black.
Now the rest of the collection consists of
4 blue
4 green
2 black
and there's space for 2 more marbles in the case.
So the question really asks: "In how many ways can 2 marbles
be selected from 4 blue ones, 4 green ones, and 2 black ones ?"
-- Well, there are 10 marbles all together.
So the first one chosen can be any one of the 10,
and for each of those,
the second one can be any one of the remaining 9 .
Total number of ways to pick 2 out of the 10 = (10 x 9) = 90 ways.
-- BUT ... there are not nearly that many different combinations
to wind up with in the case.
The first of the two picks can be any one of the 3 colors,
and for each of those,
the second pick can also be any one of the 3 colors.
So there are actually only 9 distinguishable ways (ways that
you can tell apart) to pick the last two marbles.
Answer:
Step-by-step explanation:
To make d the subject of formula, we need to rearrange the equation such that we arrive at d= _____.
<em>Remove the fraction by multiplying (d +3) on both sides:</em>
<em>Expand</em><em>:</em>
<em>
</em>
<em>Bring</em><em> </em><em>all</em><em> </em><em>the</em><em> </em><em>d</em><em> </em><em>terms</em><em> </em><em>to</em><em> </em><em>one</em><em> </em><em>side</em><em> </em><em>and</em><em> </em><em>move</em><em> </em><em>the</em><em> </em><em>others</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>other</em><em> </em><em>side</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em>:</em>
<em>Factorise</em><em> </em><em>d</em><em> </em><em>out</em><em>:</em>
<em>
</em>
<em>Divide</em><em> </em><em>by</em><em> </em><em>(</em><em>c</em><em> </em><em>+</em><em>1</em><em>)</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em>:</em>
<em></em>
