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3 years ago

The Sum of The Angles in a Triangle please dont answer yet have to add attachemnt

Mathematics

2 answers:

Delicious77 [7]3 years ago

8 0

Answer:

<POM=110°

I hope that's what you're looking for

egoroff_w [7]3 years ago

3 0

Answer:

Can I have branliest please!

The measurement is 110 degrees

Step-by-step explanation:

To find there are two ways to get it

There is a rule that states to find one of the exterior angles you have to add the other two angles that are not supplementary to the exterior one.

So, you can add 40 + 70 to get 110 degrees

You can find the missing angle in the triangle to get 70

Angle MON and Angle MOP are supplementary

so you can subtract 70 from 180 to get 110

Hope this helps!

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A package of 30 scented pencils costs $24.90. At this rate, what is the

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Answer:

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2 years ago

Can someone help me with these problems? I'm in k12 and this assignment was in classkick, so if anyone already completed this an

kherson [118]

 SOLVING QUESTIONS FROM 1ST PAGE

Given the point B(-5, 0)

y = -3x - 5

Putting x = -5, and y = 0 in y = -3x - 5

0 = -3(-5) - 5

0 = 15 - 5

0 = 10 ∵Putting B(-5, 0) in y = -3x - 5 does not equate the equation.

As L.H.S ≠ R.H.S

So,

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

Given the point C(-2, 1)

y = -3x - 5

Putting x = -2, and y = 1 in y = -3x - 5

1 = -3(-2) - 5

1 = 6 - 5

1 = 1 ∵Putting C(-2, 1) in y = -3x - 5 rightly equates the equation.

As L.H.S = R.H.S

So,

Does this check?

Answer: yes ∵ L.H.S = R.H.S

Given the point D(-1, -2)

y = -3x - 5

Putting x = -1, and y = -2 in y = -3x - 5

-2 = -3(-1) - 5

-2 = 3 - 5

-2 = -2 ∵Putting D(-1, -2) in y = -3x - 5 rightly equates the equation.

As L.H.S = R.H.S

So,

Does this check?

Answer: yes ∵ L.H.S = R.H.S

 SOLVING QUESTIONS FROM 2ND PAGE

Given the point B(6, 18)

y = -2x + 18

Putting x = 6, and y = 18 in y = -2x + 18

18 = -2(6) + 18

18 = -12 + 18

18 = 6 ∵Putting B(6, 18) in y = -2x + 18 does not equate the equation.

As L.H.S ≠ R.H.S

So,

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

Given the point C(9, 24)

y = 2x + 6

Putting x = 9, and y = 24 in y = 2x + 6

24 = 2(9) + 6

24 = 18 + 6

24 = 24 ∵Putting C(9, 24) in y = 2x + 6 rightly equates the equation.

As L.H.S = R.H.S

So,

Does this check?

Answer: yes ∵ L.H.S = R.H.S

 SOLVING QUESTIONS FROM 3RD PAGE

Given the point D(2, 7)

y = 3x + 4

Putting x = 2, and y = 7 in y = 3x + 4

7 = 3(2) + 4

7 = 6 + 4

7 = 10 ∵Putting D(2, 7) in y = 3x + 4 does not equate the equation.

As L.H.S ≠ R.H.S

So,

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

 SOLVING QUESTIONS FROM 4th PAGE

b. Which function could have produced the values in the table.

A. y = 3x + 4 

B. y = -2x + 18

C. y = 2x + 6

D. y = x + 9

The Table:

x y

3 12

6 18

9 24

Checking A) y = 3x + 4

Putting (3, 12), (6, 18) and (9, 24) in y = 3x + 4

For (3, 12)

y = 3x + 4

12 = 3(3) + 4

12 = 13 ∵ L.H.S ≠ R.H.S

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

For (6, 18)

18 = 3(6) + 4

18 = 18 + 4

18 = 22 ∵ L.H.S ≠ R.H.S

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

For (9, 24)

24 = 3(9) + 4

24 = 27 + 4

24 = 31 ∵ L.H.S ≠ R.H.S

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

So, the equation y = 3x + 4 could have not produced the all values in the table as the the ordered pairs in table do not satisfy(equate) the equation.

Checking B) y = -2x + 18

Putting (3, 12), (6, 18) and (9, 24) in y = -2x + 18

For (3, 12) ⇒ y = -2x + 18 ⇒ 12 = -2(3) + 18 ⇒ 12 = 12 ⇒ L.H.S = R.HS
For (6, 18) ⇒ y = -2x + 18 ⇒ 18 = -2(6) + 18 ⇒ 18 = 6 ⇒ L.H.S ≠ R.HS
For (9, 24) ⇒ y = -2x + 18 ⇒ 24 = -2(9) + 18 ⇒ 18 = 0 ⇒ L.H.S ≠ R.HS

So, y = -2x + 18 could have not produced all the values in the table, as (6, 18) does not equate the equation.

Checking C) y = 2x + 6

Putting (3, 12), (6, 18) and (9, 24) in y = 2x + 6

For (3, 12) ⇒ y = 2x + 6 ⇒ 12 = 2(3) + 6 ⇒ 12 = 12 ⇒ L.H.S = R.HS
For (6, 18) ⇒ y = 2x + 6 ⇒ 18 = 2(6) + 6 ⇒ 18 = 18 ⇒ L.H.S = R.HS
For (9, 24) ⇒ y = 2x + 6 ⇒ 24 = 2(9) + 6 ⇒ 24 = 24 ⇒ L.H.S = R.HS

So, y = 2x + 6 could have produced the values of in the table as all the orders pairs in the table satisfy/equate the equation.

Checking D) y = x + 9

Putting (3, 12), (6, 18) and (9, 24) in y = x + 9

For (3, 12) ⇒ y = x + 9 ⇒ 12 = 3 + 9 ⇒ 12 = 12 ⇒ L.H.S = R.HS
For (6, 18) ⇒ y = x + 9 ⇒ 18 = 6 + 9 ⇒ 18 = 15 ⇒ L.H.S ≠ R.HS
For (9, 24) ⇒ y = x + 9 ⇒ 24 = 9 + 9 ⇒ 24 = 18 ⇒ L.H.S ≠ R.HS

So, y = x + 9 could have also not produced all the values in the table, as (6, 18) and (9, 24) do not satisfy/equate the equation.

So, from all the verification we conclude that:

HENCE, ONLY y = 2x + 6 COULD HAVE PRODUCED THE VALUES IN THE TABLE AS ALL THE ORDERED PAIRS OF THE TABLE SATISFY/EQUATE THE EQUATION.

 SOLVING QUESTIONS FROM 5th PAGE

What is the domain and range of the relation?

{(-3, 7), (6, 2), (5, 1), (-9, -6)}

Domain: Domain is the set of all the x-coordinates of the ordered pairs of the relation, meaning the all first elements of the ordered pairs in a relation include in the domain of the relation.
Range: Range is the set of all the y-coordinate of the ordered pairs of the relation, meaning the all second elements of the ordered pairs in a relation include in the range of the relation.

As the given relation is {(-3, 7), (6, 2), (5, 1), (-9, -6)}

The domain is: {-3, 6, 5, -9}

Note: generally, we write the numbers in ascending order for both the domain and range.

The domain could also be written in order as: {-9, -3, 5, 6}

The range is: {7, 2, 1, -6}

The range could also be written in order as: {-6, 1, 2, 7}

Keywords: equation, point, ordered pair, domain, range

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You didn'tlist any but anyway

$x^{-m}=\frac{1}{x^m}$
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