First set everything to 0.
Giving you A= 1 , B= 1 and C= -5
Then input everything into the quadratic formula and either solve by hand or use a calculator.
Note exact values are found by hand, decimals are found using a calculator.
<span>Solve for each equation
h divided by 4/9 for h = 5 1/3
h 16/3
------- = -----------------
4/9 4/9
= 16/3 * 9/4
= 12
answer is </span><span>C) 12</span>
10 x 10 = 100
100 x 5 = 500
10 x 10 x 10 = 1,000
1,000 x 4 = 4,000
4,000 + 500 = 4,500
Hope this helped!
Answer:
side b= 11.489
Step-by-step explanation:
well I'm doing the exact same lesson currently and I just did the test on this exact question soo yeahh
-- 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.
