What is the solution to 0.4(12 – 3x) = 0.3(12x - 16)?
1 answer:
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Answer:
(i) (x +4)^3 = x^3 +12x^2 +48x +64
(ii) (2m +1)^3 = 8m^3 +12m^2 +6m +1
(iii) (a -2b)^3 = a^3 -6a^2b +12ab^2 -8b^3
Step-by-step explanation:
Powers of a binomial have a pattern. The coefficients of the terms of the expanded product are the numbers in the corresponding row of Pascal's Triangle. Here, you want the 3rd power, so the coefficients of interest are ...
1, 3, 3, 1
The pattern is ...
(a +b)^2 = a^2 +2ab +b^2
(a +b)^3 = a^3 +3a^2b +3ab^2 +b^3
The total of powers of the variable is equal to the exponent of the binomial. The powers of the first variable decrease to 0 while the powers of the second variable increase from zero.
When 'a' or 'b' is something with a coefficient, that whole term is raised to the appropriate power.
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i) (x+4)^3 = x^3 +3·4x^2 +3·4^2x +4^3
(x +4)^3 = x^3 +12x^2 +48x +64
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ii) (2m +1)^3 = (2m)^3 +3·1·(2m)^2 +3·1^2·(2m) +1^3
(2m +1)^3 = 8m^3 +12m^2 +6m +1
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iii) (a -2b)^3 = a^3 +3·a^2·(-2b) +3·a·(-2b)^2 +(-2b)^3
(a -2b)^3 = a^3 -6a^2b +12ab^2 -8b^3
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<em>Additional comment on Pascal's triangle</em>
The numbers in each row are the sums of the two numbers immediately above. For binomial expansion, you're only interested in the contents of a row. The "patterns" shown in the attachment are an extra added attraction, not particularly relevant to binomial expansion.
The firts thig we are going to do is create tow triangles using the angles of elevation of Paul and Jose. Since the problem is not giving us their height we'll assume that the horizontal line of sight of both of them coincide with the base of the tree.
We know that Paul is 19m from the base of the tree and its elevation angle to the top of the tree is 59°. We also know that the elevation angle of Jose and the top of the tree is 43°, but we don't know the distance between Paul of Jose, so lets label that distance as
.
Now we can build a right triangle between Paul and the tree and another one between Jose and the tree as shown in the figure. Lets use cosine to find h in Paul's trianlge:
Now we can use the law of sines to find the distance between Paul and Jose:
Now that we know the distance between Paul and Jose, the only thing left is add that distance to the distance from Paul and the base of the tree:
We can conclude that Jose is 33.9m from the base of the tree.
We know that Paul is 19m from the base of the tree and its elevation angle to the top of the tree is 59°. We also know that the elevation angle of Jose and the top of the tree is 43°, but we don't know the distance between Paul of Jose, so lets label that distance as
Now we can build a right triangle between Paul and the tree and another one between Jose and the tree as shown in the figure. Lets use cosine to find h in Paul's trianlge:
Now we can use the law of sines to find the distance
Now that we know the distance between Paul and Jose, the only thing left is add that distance to the distance from Paul and the base of the tree:
We can conclude that Jose is 33.9m from the base of the tree.