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3 years ago

7

Can some one help me with this

2 answers:

Fittoniya [83]3 years ago

8 0

The answer is <u>A, parallel</u>. The slopes are the same (<em>m=-4/5</em>)

I hope this helped :D

anyanavicka [17]3 years ago

3 0

Its parallel for this math problem

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4y + 4 = 2y - 10 <br> can somebody break this down to solve it ? please

vredina [299]

Answer:

y=-7

Step-by-step explanation:

first, you want variables on one side, numbers on the other side of the equation.

4y+4=2y-10

*plus ten on both side to get rid of -10*

4y+14=2y

*minus 4y on each side to get rid of 4y*

14=-2y

*divide by -2 on each side to get the value of just one y itself*

y=-7

8 0

3 years ago

What is 411, 600, 000 in scientific notation

Morgarella [4.7K]

Answer:

4.116 × 108

Step-by-step explanation:

I simply used a Scientific Notation Converter

5 0

3 years ago

Read 2 more answers

What is the relationship between 0.04 and 0.004

olga_2 [115]

0.04 is ten times more than 0.004

4 0

3 years ago

determine if the ordered pair is a solution of the equation is (2,4) a solution of y = 10 -3×? true or false

anzhelika [568]

Answer:

False

Step-by-step explanation:

4 = 10 - 3²

4 = 10 - 9

4 = 1

7 0

3 years ago

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

3 years ago