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3 years ago

10

HELPPPPPPPP LOL I NEED HELPPPPPPPP

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lorasvet [3.4K]3 years ago

7 0

Answer:

<h3>2nd row is wrong b</h3><h2>sat , st , ts </h2>

<h3>hope you got your answer here </h3><h2>please give brainliest plz follow </h2>

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Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos

Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$P_{1} = (- 1, 4)$

$P_{2} = (6, 2)$ and

$P_{3} = (4, - 5)$

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$

Length between $P_1$ and $P_{2}$ , (Side 1) :

$(x_1, y_1) = (- 1, 4)$ and

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2$

$= \sqrt{7^2 + 2^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}}$

Length of Side 1 = $\sqrt{\textbf{53}}$ units.

Distance between $P_1$ and $P_2$ , (Side 2):

$(x_1, y_1) = (-1, 4)$

$(x_2, y_2) = (4, - 5)$

Distance = $\sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2$

$= \sqrt{ 25 + 81 }$

$= \sqrt{\textbf{106}}$

Length of Side 2 = $\sqrt{\textbf{106}}$ units.

Distance between $P_2$ and $P_3$ , Side 3 :

$(x_1, y_1) = (4, 5)$

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}$

Length of Side 3 = $\sqrt{\textbf{53}}$ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

$\bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2$

i.e, 53 + 53 = 106

Hence, the triangle is a right - angled triangle as well.

7 0

4 years ago

Fran need to move $9 million out of the country. The dimensions of a US bill is 2.61 inches by 6.14 inches by 0.0043 inches.

Mice21 [21]

Answer:

1728 in³ in a cubic ft
3.6 cubic feet (rounded to tenths)
2 suitcases

Step-by-step explanation:

For part 2, you have rounded to hundredths. The question asks for the answer rounded to tenths.

__

<u>Part 3</u>

(3.6 ft³)×(1 suitcase)/(2.9 ft³) = 3.6/2.9 suitcases ≈ 1.24 suitcases

Suitcases only come in whole numbers. 1 is too few, so 2 suitcases are needed.

5 0

3 years ago

Find side A of a triangle with a B length of 7 and a C length of 25

Vikentia [17]

Answer:

A 24

Step-by-step explanation:

Assuming this is a right triangle, we can use the Pythagorean theorem

a^2+ b^2 = c^2

a^2 + 7^2 = 25^2

a^2 +49= 625

Subtract 49 from each side

a^2 +49-49 = 625-29

a^2 = 576

Take the square root of each side

sqrt(a^2) = sqrt(576)

a = 24

4 0

3 years ago

Read 2 more answers

The height, h, in feet of a ball suspended from a spring as a function of time, t, in seconds can be modeled by the equation (eq

uysha [10]

Answer:

D

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

3x3-27x=0<br> Solve by factoring a

Eddi Din [679]

After factoring $3x^3-27x=0$ the solutions are x=0 , x=3, x= -3

Step-by-step explanation:

We need to solve by factoring:

$3x^3-27x=0$

Solving:

$3x^3-27x=0$

Taking 3x common:

$3x(x^2-9)=0$

Now using formula $a^2-b^2=(a-b)(a+b)$

$3x((x)^2-(3)^2)=0\\3x(x-3)(x+3)=0$

$3x=0\,\,and\,\,x+3=0\,\,and\,\,x-3=0\\x=0\,\,and\,\,x=-3\,\,and\,\,x=3\\$

So, After factoring $3x^3-27x=0$ the solutions are x=0 , x=3, x= -3

Keywords: Factorization

Learn more about Factorization at:

#learnwithBrainly

5 0

3 years ago