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2 years ago

Given the vectors, a=3i+4j, b=-2i+5j, c=10i-j, d=-1/3i+5/2j, find -0.4a-0.3b+0.2d=?

Mathematics

1 answer:

sveta [45]2 years ago

6 0

Answer:

$-\frac{2}{3}i - \frac{13}{5}j$

Step-by-step explanation:

-0.4(3i + 4i) - 0.3(-2i + 5j) + 0.2(- $\frac{1}{3}i$ + $\frac{5}{2}$ j)

-1.2i - 1.6j + 0.6i - 1.5j - $\frac{0.2}{3}$ i + 0.5j

Converting to fraction form;

$-\frac{6}{5}i + \frac{3}{5}i - \frac{1}{15}i - \frac{8}{5}j - \frac{3}{2}j + \frac{1}{2} j$

<u>Solving the i part;</u>

$\frac{-18i + 9i - i}{15}$ = $-\frac{2}{3}i$

<u>Solving the j part;</u>

$\frac{-16j-15j+5j}{10} = -\frac{13}{5}$ j

So -0.4a - o.3b + 0.2d = $-\frac{2}{3}i$ - $\frac{13}{5}j$

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2 years ago

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Answer:

Therefore $k= \frac{ln2 }{18}$ , A=184

Step-by-step explanation:

Given function is

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2 years ago

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