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hammer [34]
2 years ago
14

In quadrilateral PQRS, measures (7x - 2)o . Angle PSR measures (5x + 14 )o. What are the measure of angles PQR and PSR? m = 54o

and m = 54o m = 84o and m = 96o m = 90o and m = 90o m = 96o and m = 84o
Mathematics
2 answers:
True [87]2 years ago
8 0

Answer:

Step-by-step explanation:

Given that PQRS is a parallelogram and angles PSR = 5x+14 and QPS = 7x-2

Since these two are adjacent angles, the sum will be equal to 180

5x+14+7x-2=180

12x=180-12 =168

x=14

Substitute 14 for x to get

QPS= 7x-2 = 98-2 =96

and PSR = 5x+14 = 70+14 = 84

Verify:

Addition of these two come to 180, proving they are supplementary

Hence verified

fiasKO [112]2 years ago
8 0

The correct answer is:

D. m ∠PQR = 96° and m ∠PSR = 84°

The measure of angles PQR and PSR is 96° and 84°.

|Huntrw6|

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Answer:

3rd graph down

Step-by-step explanation:

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2 years ago
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Mrs.Steffen’s third grade class has 30 students in it. The students are divided into three groups(numbered 1, 2,and 3),each havin
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Answer:

a. \\ 10! = 3628800;

b. \\ 10!*10!*10! = 47784725839872000000 = 4.7784725839872*10^{19}

Step-by-step explanation:

We need here to apply the <em>Multiplication Principle </em>or the <em>Fundamental Principle of Counting</em> for each answer. Answer <em>b</em> needs an extra reasoning for being completed.

The <em>Multiplication Principle</em> states that if there are <em>n</em> ways of doing something and <em>m</em> ways of doing another thing, then there are <em>n</em> x <em>m</em> ways of doing both (<em>Rule of product</em> (2020), in Wikipedia).

<h3>In how many ways can ten students line up? </h3>

There are <em>ten</em> students. When one is selected, there is no other way to select it again. So, <em>no repetition</em> is allowed.

Then, in the beginning, there are 10 possibilities for 10 students; when one is selected, there are nine possibilities left. When another is selected, eight possibilities are left to form the file, and so on.

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This could be expressed mathematically using n!:

\\ n! = n * (n-1)! * (n-2)! *...* 2*1.

For instance, \\ 5! = 5 * (5-1)! * (5-2)! *...*2*1 = 5 * 4 * 3 * 2 * 1 = 120.

So, for the case in question, the <em>ten</em> students can line up in:

\\ 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800 ways to line up in a single file.

<h3>Second Question</h3>

For this question, we need to consider the former reasoning with extra consideration in mind.

The members of Group 1 can occupy <em>only</em> the following places in forming the file:

\\ G1 = \{ 1, 4, 7, 10, 13, 16, 19, 22, 25, 28\}^{th} <em>places</em>.

The members of Group 2 <em>only</em>:

\\ G2 = \{ 2, 5, 8, 11, 14, 17, 20, 23, 26, 29\}^{th} <em>places</em>.

And the members of Group 3, the following <em>only</em> ones:

\\ G3 = \{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}^{th} <em>places.</em>

Well, having into account these possible places for each member of G1, G2 and G3, there are: <em>10! ways</em> for lining up members of G1; <em>10! ways</em> for lining up members of G2 and, also, <em>10! ways</em> for lining up members of G3.

After using the <em>Multiplication Principle</em>, we have, thus:

\\ 10! * 10! * 10! = 47784725839872000000 = 4.7784725839872 *10^{19} <em>ways the students can line up to come in from recess</em>.

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Answer:

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