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fredd [130]
3 years ago
10

Math: Please help me I don’t understand and I need to do corrections.

Mathematics
1 answer:
GREYUIT [131]3 years ago
4 0
So, you had done everything right so far (other than squaring the 2), but that was only half of the question.

to find the least common multiple, you need to first figure out what the prime factors have in common.
{2}^{2}  \times 3 \times 5 \\ and \\  {2}^{2}  \times  {3}^{2}  \times 5 \times 7
each have two twos. both have one 5, so we know our answer will look something like
{2}^{2}  \times 5 \times other \: stuff
now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.

one has one 3 and one has two, so we need two threes. now our equation is
{2}^{2}  \times {3}^{2}  \times 5 \times stuff

what's the only number we have to deal with? 7...

how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes
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Suppose that f(x) is an EVEN function and let the integral f(x) dx from 0 to 1=5 and the integral f(x) dx from 0 to 7 =1. What i
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<span>Since it's even, the integral from -7 to -1 is the same as the integral from 1 to 7.
</span><span>think about
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</span>
<span>from -7..-1 of x^2 its -1/3+7^3/3 to get this without integrating we just do 7^3/3-1/3 which is the integral from 0..7 minus the integral from 0..1
</span><span>So the integral from 1 to 0 f(x) dx = 5 changes to the integral from 0 to 1 f(x) dx = -5 and you add it to the integral from 0 to 7 which is 1 making -4</span>
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