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3 years ago

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Triangle ABC has vertices A (3,-3) B (-7,9) C (-11,3). Determine the point of intersection of the medians, and state the coordin

ates

1 answer:

emmainna [20.7K]3 years ago

5 0

Answer:

(-5, 3)

Step-by-step explanation:

Let O(x, y) be the centroid of the triangle.

The median of a triangle is a line segment that joins the vertex of a triangle to the midpoint of the opposite side. There are three medians in a triangle; and the medians of a triangle intersect at a point called the centroid.

The centroid of a triangle is gotten from the average of the x coordinates and the y coordinates of all the three vertices.

For Triangle ABC has vertices A (3,-3) B (-7,9) C (-11,3) and centroid O(x, y). Hence:

$x=\frac{3 + (-7)+(-11)}{3}=-5\\\\y=\frac{-3 + 9 + 3}{3} =3$

The centroid is at (-5, 3)

You might be interested in

Use the Polynomial Identity below to help you create a list of 10 Pythagorean Triples:

Stolb23 [73]

Answer:

(3,4,5)

(6,8,10)

(5,12,13)

(8,15,17)

(12,16,20)

(7,24,25)

(10,24,26)

(20,21,29)

(16,30,34)

(9,40,41)

Just choose 2 numbers from {1,2,3,4,5,6,7,8,...} and make sure the one you input for x is larger.

Post the three in the comments and I will check them for you.

Step-by-step explanation:

We need to choose 2 positive integers for x and y where x>y.

Positive integers are {1,2,3,4,5,6,7,.....}.

I'm going to start with (x,y)=(2,1).

x=2 and y=1.

$(2^2+1^2)^2=(2^2-1^2)^2+(2\cdot2\cdot1)^2$

$(4+1)^2=(4-1)^2+(4)^2$

$(5)^2=(3)^2+(4)^2$

So one Pythagorean Triple is (3,4,5).

I'm going to choose (x,y)=(3,1).

x=3 and y=1.

$(3^2+1^2)^2=(3^2-1^2)^2+(2\cdot3\cdot1)^2$

$(9+1)^2=(9-1)^2+(6)^2$

$(10)^2=(8)^2+(6)^2$

So another Pythagorean Triple is (6,8,10).

I'm going to choose (x,y)=(3,2).

x=3 and y=2.

$(3^2+2^2)^2=(3^2-2^2)^2+(2\cdot3\cdot2)^2$

$(9+4)^2=(9-4)^2+(12)^2$

$(13)^2=(5)^2+(12)^2$

So another is (5,12,13).

I'm going to choose (x,y)=(4,1).

$(4^2+1^2)^2=(4^2-1^2)^2+(2\cdot4\cdot1)^2$

$(16+1)^2=(16-1)^2+(8)^2$

$(17)^2=(15)^2+(8)^2$

Another is (8,15,17).

I'm going to choose (x,y)=(4,2).

$(4^2+2^2)^2=(4^2-2^2)^2+(2\cdot4\cdot2)^2$

$(16+4)^2=(16-4)^2+(16)^2$

$(20)^2=(12)^2+(16)^2$

We have another which is (12,16,20).

I'm going to choose (x,y)=(4,3).

$(4^2+3^2)^2=(4^2-3^2)^2+(2\cdot4\cdot3)^2$

$(16+9)^2=(16-9)^2+(24)^2$

$(25)^2=(7)^2+(24)^2$

We have another is (7,24,25).

You are just choosing numbers from the positive integer set {1,2,3,4,... } and making sure the number you plug in for x is higher than the number for y.

I will do one more.

Let's choose (x,y)=(5,1).

$(5^2+1^2)^2=(5^2-1^2)^2+(2\cdot5\cdot1)^2$

$(25+1)^2=(25-1)^2+(10)^2$

$(26)^2=(24)^2+(10)^2$

So (10,24,26) is another.

Let (x,y)=(5,2).

$(5^2+2^2)^2=(5^2-2^2)^2+(2\cdot5\cdot2)^2$

$(25+4)^2=(25-4)^2+(20)^2$

$(29)^2=(21)^2+(20)^2$

So another Pythagorean Triple is (20,21,29).

Choose (x,y)=(5,3).

$(5^2+3^2)^2=(5^2-3^2)^2+(2\cdot5\cdot3)^2$

$(25+9)^2=(25-9)^2+(30)^2$

$(34)^2=(16)^2+(30)^2$

Another Pythagorean Triple is (16,30,34).

Let (x,y)=(5,4)

$(5^2+4^2)^2=(5^2-4^2)^2+(2\cdot5\cdot4)^2$

$(25+16)^2=(25-16)^2+(40)^2$

$(41)^2=(9)^2+(40)^2$

Another is (9,40,41).

5 0

3 years ago

Help FAST PLZ!!! marking brainliest!

ss7ja [257]

Answer:

After finding the prime factorization of $2010=2\cdot3\cdot5\cdot67$, divide $5300$ by $67$ and add $5300$ divided by $67^2$ in order to find the total number of multiples of $67$ between $2$ and $5300$. $\lfloor\frac{5300}{67}\rfloor+\lfloor\frac{5300}{67^2}\rfloor=80$ Since $71$,$73$, and $79$ are prime numbers greater than $67$ and less than or equal to $80$, subtract $3$ from $80$ to get the answer $80-3=\boxed{77}\Rightarrow\boxed{D}$.

Step-by-step explanation:

hope this helps

5 0

3 years ago

Cameron sponsor a fun raiser that 582 people attended he raise $6480 he charged $15 for balcony seats and $10 for crown seats ho

bagirrra123 [75]

Answer:

132 people bought balcony tickets and 450 bout ground tickets

Step-by-step explanation:

let x be balcony and y for ground

x+y=582

15x+10y=6480 solve by adding(subtracting)/eliminating process

multiply first equation by 15 to eliminate x:

15x+15y=8730

15x+10y=6480 subtract

15x+15y-15x-10y=8730-6480

5y=2250

y=2250/5=450

x+y=582

x=582-450=132

8 0

3 years ago

If 7x is an odd integer, represent the next odd integer as an algebraic expression in x.

blagie [28]

Given :

7x is an odd integer.

To Find :

The next odd integer as an algebraic expression in x.

Solution :

All positive integers are :

1 , 2 , 3 , 4 , 5 , ......

From the above series we noticed that every other odd number is 2 more than the previous one.

So , 7x is an odd number.

Therefore, next odd number is 7x + 2.

Hence, this is the required solution.

3 0

3 years ago

We know that a triangle with side lengths , , and is a right triangle. Using those side lengths, find the missing triples and x-

Anit [1.1K]

The completed table for the Pythagorean triples are:

$x \text{ value } \text{ Pythagorean triple}\\\text{ } \text{ } \text{ } 3 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (6,8,10)\\\text{ } \text{ } \text{ } 4 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (8,15,17)$

$\text{ } \text{ } \text{ } 5 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (10,24,26)\\\text{ } \text{ } \text{ } 6 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (12,35,37)$ .

A trio of positive numbers known as a Pythagorean triple fits into the Pythagoras theorem's formula, which is written as a² + b² = c², where a, b, and c are all positive integers. Here, "a" and "b" make up the other two legs of the right-angled triangle, and "c" serves as the triangle's "hypotenuse," or longest side. In terms of the Pythagorean triples (a, b, c).

In the question, we are given that a triangle with side lengths x² - 1, 2x, x² + 1 is a right triangle, and are asked to fill in the missing table:

$x \text{ value } \text{ Pythagorean triple}\\\text{ } \text{ } \text{ } 3 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \underline{\text{ }}\\\text{ } \text{ } \text{ } \underline{\text{ }} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (8,15,17)\\$

$\text{ } \text{ } \text{ } 5 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \underline{\text{ }}\\\text{ } \text{ } \text{ } \underline{\text{ }} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (12,35,37)$ .

Taking x as 3, and putting in the sides, we get:

x² + 1 = 3² + 1 = 10.

2x = 2*3 = 6.

x² - 1 = 3² - 1 = 8.

Thus, the triplet is 6,8,10.

For the triplet 8,15,17, we get the x value as:

2x = 8,

or, x = 8/2 = 4.

Thus, the x-value for the triplet (8,15,17) is 4.

Taking x as 5, and putting in the sides, we get:

x² + 1 = 5² + 1 = 26.

2x = 2*5 = 10.

x² - 1 = 5² - 1 = 24.

Thus, the triplet is 10,24,26.

For the triplet 12,35,37, we get the x value as:

2x = 12,

or, x = 12/2 = 6.

Thus, the x-value for the triplet (12,35,37) is 6.

Thus, the completed table for the Pythagorean triples are:

$x \text{ value } \text{ Pythagorean triple}\\\text{ } \text{ } \text{ } 3 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (6,8,10)\\\text{ } \text{ } \text{ } 4 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (8,15,17)$

$\text{ } \text{ } \text{ } 5 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (10,24,26)\\\text{ } \text{ } \text{ } 6 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (12,35,37)$ .

Learn more about the Pythagorean triplets at

brainly.com/question/24211694

#SPJ4

The provided question is incomplete. The complete question is:

"We know that a triangle with side lengths x² - 1,2x and x² + 1 is a right triangle. Using those side lengths, find the missing triples and x-values.

Write the triples in parentheses, without spaces between the numbers, and with a comma between numbers. Write the triples in order from least to greatest.

Type the correct answer in each box.

$x \text{ value } \text{ Pythagorean triple}\\\text{ } \text{ } \text{ } 3 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \underline{\text{ }}\\\text{ } \text{ } \text{ } \underline{\text{ }} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (8,15,17)\\$

$\text{ } \text{ } \text{ } 5 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \underline{\text{ }}\\\text{ } \text{ } \text{ } \underline{\text{ }} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (12,35,37)$ .

6 0

2 years ago