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2 years ago

Using the identity sin? O + cos? O = 1, find the value of tan O, to the nearest

Mathematics

1 answer:

Novosadov [1.4K]2 years ago

7 0

Answer:

3.7

Step-by-step explanation:

Given

sin²θ + cos²θ = 1
sinθ = 0.27

Solving for cosθ

cos²θ = 1 - sin²θ
cos²θ = 1 - (0.27)²
cos²θ = 0.0729

Finding tanθ

tanθ = sinθ / cosθ
tanθ = 0.27 / 0.0729
tanθ = 3.7

Verifying it lies in the range

θ = tan⁻¹ (3.7)
θ = 74.88°
Range is : 0 < θ < π/2 [or 0 < θ < 90°]
It lies in the range [Verified]

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2(x-1) ≥ 10 or 3-4x > 15 Solve the compound inequality, and show work!

Veseljchak [2.6K]

Answer:

$x\geq 6\text{ or } x$

Step-by-step explanation:

We have the compound inequality:

$2(x-1)\geq10\text{ or } 3-4x>15$

Let's solve each of them individually first:

We have:

$2(x-1)\geq10$

Divide both sides by 2:

$x-1\geq5$

Add 1 to both sides:

$x\geq6$

We have:

$3-4x>15$

Subtract from both sides:

$-4x>12$

Divide both sides by -4:

$x$

Hence, our solution set is:

$x\geq 6\text{ or } x$

5 0

2 years ago

Read 2 more answers

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

(c) The novels must be together and the chemistry books, together

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$