Part I)
The module of vector AB is given by:
lABl = root ((- 3) ^ 2 + (4) ^ 2)
lABl = root (9 + 16)
lABl = root (25)
lABl = 5
Part (ii)
The module of the EF vector is given by:
lEFl = root ((5) ^ 2 + (e) ^ 2)
We have to:
lEFl = 3lABl
Thus:
root ((5) ^ 2 + (e) ^ 2) = 3 * (5)
root ((5) ^ 2 + (e) ^ 2) = 15
Clearing e have:
(5) ^ 2 + (e) ^ 2 = 15 ^ 2
(e) ^ 2 = 15 ^ 2 - 5 ^ 2
e = root (200)
e = root (2 * 100)
e = 10 * root (2)
Well, that means that it is 3 more than a common multipule of 8,10,12, and 30
find the lcm
8=2*2*2
10=2*5
12=2*2*3
30=2*3*5
lcm=2*2*2*3*5=120
times it by 2 to get it between 200 and 500
240
add 3
243 is da number
Answer:
Step-by-step explanation:
I think x=-2
Answer:
7
Step-by-step explanation:
The order of operations tells you to start any evaluation by looking at the innermost set of parentheses first.
Here, that means your first step is to find the value of h(-3). You do that by finding the input (x) value -3 in the table for h(x), and locating the corresponding output, h(x), which is 2.
Now, the problem becomes evaluating g(2).
You do the same thing for that function: locate the input x=2 in the table for g(x) and find the corresponding output: 7.
Now, you know ...
g(h(-3)) = g(2) = 7
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