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3 years ago

Need help with this problem

Mathematics

2 answers:

bekas [8.4K]3 years ago

7 0

Answer:

Step-by-step explanation:

In this problem, it is given that lines (AO) and (OC) are perpendicular. This means that when the lines intersect, the four angles formed each have a measure of (90) degrees. As per the given diagram, angles (<AOR) and (<ROC) add up to make (<AOC). Thus, one can form the following equation and solve for (x).

$($

Substitute,

$(6x-6)+(5x+8)=90$

Simplify,

$11x+2=90$

Inverse operations,

$11x+2=90$

$11x=88\\\\x=8$

Substitute this into the equation for the measure of angle (<BOC) to find the numerical angle measure:

e-lub [12.9K]3 years ago

4 0

Answer:

OA⊥OC

∴∠AOB+BOC= 90°

5x+8+6n-6=90

11x+2=90

x=90-2/11

x=88/11

x=8

∠BOC=6x-6

=6(8)-6

=48-6

∠BOC=42°

------------------------------

Hope it helps...

Have a great day!!

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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
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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:

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

$\mathbf{n =4}$

So, the number of arrangements is:

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

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

$\mathbf{Arrangements = 17280}$

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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:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

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So, the number of arrangements of these books is:

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

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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:

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

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

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Read more about permutations at:

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