( a ) The system of equations:
2 r + b = 8.403 r + b = 9.35( b ) Graph is in the attachment.
( c ) Each item costs:
b = $6.50, r = $0.95We can prove it: 2 * 0.95 + 6.50 = 8.40
3* 0.95 + 6.50 = 9.35
Answer:
<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>
Step-by-step explanation:
Given the function j(x) = 11.6
and k(x) =
, to show that both equality functions are true, all we need to show is that both j(k(x)) and k(j(x)) equal x,
For j(k(x));
j(k(x)) = j[(ln x/11.6)]
j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}
j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)
j[(ln x/11.6)] = 11.6 * x/11.6
j[(ln x/11.6)] = x
Hence j[k(x)] = x
Similarly for k[j(x)];
k[j(x)] = k[11.6e^x]
k[11.6e^x] = ln (11.6e^x/11.6)
k[11.6e^x] = ln(e^x)
exponential function will cancel out the natural logarithm leaving x
k[11.6e^x] = x
Hence k[j(x)] = x
From the calculations above, it can be seen that j[k(x)] = k[j(x)] = x, this shows that the functions j(x) = 11.6
and k(x) =
are inverse functions.
1.55 / 1 = x / 3.5....$ 1.55 to 1 lb = $ x to 3.5 lbs
first one <==
it could have also been : 1 / 1.55 = 3.5 / x...but that is not an answer choice
Question Completion:
How are the percentages distributed? Is the distribution skewed? Are there any gaps? (Select all that apply.)
Answer:
1. The percentages are concentrated from 20% to 60%.
2. These data are strongly skewed left.
3. There are no gaps in the data.
Step-by-step explanation:
1. Data
Percentage loss of wetlands per state
46 37 36 42 81 20 73 59 35 50
87 52 24 27 38 56 39 74 56 31
27 91 46 9 54 52 30 33 28 35
35 23 90 72 85 42 59 50 49
48 38 60 46 87 50 89 49 67
2. Re-arrangement of
Percentage loss of wetlands per state (in ascending order)
9 20 23 24 27 27 28 30
31 33 35 35 35 36 37 38
38 39 42 42 46 46 46 48
49 49 50 50 50 52 52 54
56 56 59 59 60 67 72 73
74 81 85 87 87 89 90 91
To solve this problem you must apply the proccedure shown below:
1. You have to r<span>ewrite x=12 in polar form. Then, you have:
12=rCos</span><span>θ
2. Then, you must solve for r, as following:
r=12/Cos</span><span>θ
</span> 3. You have that 1/Cosθ=Sec<span>θ, therefore:
</span> r=12(1/Cos<span>θ)
</span> r=12Sec<span>θ
</span> Therefore, as you can see, the answer is: r=12Secθ<span>
</span>