Answer:
y= -2x-22.
Step-by-step explanation:
1) the slope-interception common form is y=s*x+i, where 's' and 'i' are the slope and interception, unknown numbers;
2) if x₁= -3; y₁= -16, and s=-2, then the equation of the required line can be written in the point-interception form y-y₁=s(x-x₁); ⇔ y+16= -2(x+3);
3) the required equation in slope-interception form is:
y+16= -2x-6; ⇒ y= -2x-22.
note, the provided solution is not the only and shortest way.
<span>2 ln 8 + 3 ln y simplifies to ln 8^2 + ln y^3, which in turn simplifies to
ln 64/y^3, or ln {64*y^3}</span>
-- 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.
I agree with there answer ^
Answer:
Step-by-step explanation:
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