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3 years ago

The sum of the measures of two exterior angles of a triangle is 252. What is the measure of the third exterior angle?

Mathematics

1 answer:

alisha [4.7K]3 years ago

3 0

108 is your answer to this question

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X^2-4x+3 factorize into its factors and give me answer I will giv u anda burger

Ghella [55]

Answer:

(x - 3)(x - 1)

Step-by-step explanation:

Consider the factors of the constant term (+ 3) which sum to give the coefficient of the x- term.

The factors are - 3 and - 1, since

- 3 × - 1 = + 3 and - 3 - 1 = - 4, hence

x² - 4x + 3 = (x - 3)(x - 1) ← in factored form

6 0

3 years ago

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A circle has a diameter of 8 cm. An angle in the center of the circle creates an arc AB. That angle's measure is pi/4 rad. What

zhenek [66]

Answer:

3.14 cm

Step-by-step explanation:

From the question,

Applying

s = πdФ/360................... Equation 1

Where s = length of arc AB, d = diameter of the circle, Ф = angle at the center of the circle, π = pie

From the question,

Given: d = 8 cm, Ф = π/4 rad = 45°

Constant: π = 22/7

Substitute these values into equation 1

s = (22/7)(8)(45)/360

s = 3.14 cm.

Hence the lenght of arc AB = 3.14 cm

8 0

3 years ago

Carlos and his family left for the amsuement park at 8:35

ad-work [718]

10:30 a.m. I think

hope that helps :)

8 0

3 years ago

In a small chess tournament, 20 matches were played. Find out how many people were involved if it is known that each participant

Nina [5.8K]

Answer:

5 players participated in the tournament.

Step-by-step explanation:

In a small chess tournament, 20 matches were played.

Let us assume that n number of players participated in the tournament

As in each game 2 players play, so the number of ways they can play is,

$=\ ^nC_2$

As they played 2 games with every other participant in the tournament.

So the total number of games is,

$=\ 2\times ^nC_2$

But it is given to be 20, so

$\Rightarrow \ 2\times ^nC_2=20$

$\Rightarrow \ ^nC_2=10$

$\Rightarrow \dfrac{n!}{2!(n-2)!}=10$

$\Rightarrow \dfrac{n(n-1)}{2}=10$

$\Rightarrow {n(n-1)=20$

As $5\times 4=20$ , so we get n=5.

Therefore, 5 players participated in the tournament.

5 0

3 years ago

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The volume of a beack ball is 905 square centimeters. Find the radius of the ball to the nearest whole number.

lubasha [3.4K]

<h3>given:</h3>

$v = 905 \: {cm}^{3}$

the radius of the given ball (sphere).

<h3>solution:</h3>

$r = \sqrt[3]{ \frac{3v}{4\pi} }$

$r = \sqrt[3]{ \frac{3 \times 905}{4\pi} }$

$r = 6 \: cm$

therefore, the radius of the ball is 6 cm.

note: refer to the picture I added on how you can change r as the subject of the formula.

8 0

2 years ago

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