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nika2105 [10]
3 years ago
9

PLEASE HELP WITH THE QUESTION BELOW ASAP!! THANKS SO MUCH!

Mathematics
1 answer:
Lilit [14]3 years ago
3 0

Answer:

y = 6x - 40

Step-by-step explanation:

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Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3
omeli [17]
\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}

=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1

=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R
3 0
3 years ago
If 5\9 of a tank can be filled in 2 minutes, how many minutes will it take to fill the whole tank?
Blizzard [7]

Answer:

It would take 3.6 mins.

Step-by-step explanation:

In order to find this, we simply divide the amount of time by the amount of the tank that is filled.

2 ÷ 5/9 = 3.6

7 0
2 years ago
A boy's mother was 32 years old when he was born. Let a represent the boy's age and m represent the mother's age.
LenKa [72]

see the picture attached

5 0
2 years ago
If f (x) = 3x + 5/x, what is f(a+ 2)?
Arlecino [84]
F(a+2) = 3(a+2)+5/(a+2)
f(a+2) = 3a+6 + 5/(a+2)
8 0
3 years ago
Use the geometric probability distribution to solve the following problem. On the leeward side of the island of Oahu, in a small
Greeley [361]

Answer:

(a) \text{P(n)} = \text{p} \times \text{(1 - p)}^{n-1} ; \text{ n} = 1, 2, 3,....

(b) P(1) = 0.740, P(2) = 0.192, and P(3) = 0.050.

(c) The probability that n ≥ 4 is 0.018.

(d) The expected number of residents in the village you must meet before you encounter the first person of Hawaiian ancestry is 3.

Step-by-step explanation:

We are given that on the leeward side of the island of Oahu, in a small village, about 74% of the residents are of Hawaiian ancestry.

Let n = 1, 2, 3, … represent the number of people you must meet until you encounter the first person of Hawaiian ancestry in the village.

(a) We can observe that the above situation can be represented through the geometric distribution because the geometric distribution states that we will keep on going with the trials until we achieve our first success,

Here also, n represent the number of people you must meet until you encounter the first person of Hawaiian ancestry in the village.

So, the probability distribution of the geometric distribution is given by;

\text{P(n)} = \text{p} \times \text{(1 - p)}^{n-1} ; \text{ n} = 1, 2, 3,....

where, p = probability that the residents are of Hawaiian ancestry = 74%

(b) The probabilities that n = 1, n = 2, and n = 3 is given by;

\text{P(n)} = \text{p} \times \text{(1 - p)}^{n-1}

P(1) = \text{0.74} \times \text{(1 - 0.74)}^{1-1} = 0.74

P(2) = \text{0.74} \times \text{(1 - 0.74)}^{2-1} = 0.192

P(3) = \text{0.74} \times \text{(1 - 0.74)}^{3-1} = 0.050

(c) The probability that n ≥ 4 is given by = P(n ≥ 4)

     P(n ≥ 4) = 1 - P(n = 1) - P(n = 2) - P(n = 3)

                   = 1 - 0.74 - 0.192 - 0.050

                   = 0.018

(d) The expected number of residents in the village you must meet before you encounter the first person of Hawaiian ancestry is given by = E(n)

We know that the mean of the geometric distribution is given by;

           Mean = \dfrac{p}{1-p} = \dfrac{0.74}{1-0.74}

                     = \dfrac{0.74}{0.26} = 2.85 or 3 (approx).

4 0
3 years ago
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