Find x for brainiest 3

What is the perimeter of the triangle shown on the coordinate plane,to the nearest tenth of a unit ?

ollegr [7]

Answer:

25.6 units

Step-by-step explanation:

From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).

First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:

$d=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2}}$

where

$(x_1,y_1)$ are the coordinates of the first point

$(x_2,y_2)$ are the coordinates of the second point

- For AB:

$d=\sqrt{[1-(-5)]^{2}+(4-4)^2}$

$d=\sqrt{(1+5)^{2}+(0)^2}$

$d=\sqrt{(6)^{2}}$

$d=6$

- For BC:

$d=\sqrt{(3-1)^{2} +(-4-4)^{2}}$

$d=\sqrt{(2)^{2} +(-8)^{2}}$

$d=\sqrt{4+64}$

$d=\sqrt{68}$

$d=8.24$

- For AC:

$d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}$

$d=\sqrt{(3+5)^{2} +(-8)^{2}}$

$d=\sqrt{(8)^{2} +64}$

$d=\sqrt{64+64}$

$d=\sqrt{128}$

$d=11.31$

Next, now that we have our lengths, we can add them to find the perimeter of our triangle:

$p=AB+BC+AC$

$p=6+8.24+11.31$

$p=25.55$

$p=25.6$

We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.

6 0

3 years ago

Which expression gives the distance between the points (2,4) and (-4,8)?

cluponka [151]

(y2-y1)/(x2-x1)=(8-4)/(-4-2)

4 0

3 years ago

Read 2 more answers

HEY CAN ANYONE PLS ANSWER DIS MATH QUESTION!!!!

tangare [24]

Answer: 8

Step-by-step explanation:

7 0

2 years ago

Find the least mean square estimate ofxbased on the observationy=y.

bija089 [108]

Ndenkdendekwnndnndnd

7 0

3 years ago

Indicate the equation of the given line in standard form. Show all your work for full credit. the line containing the median of

alukav5142 [94]

Answer:

* The equation of the median of the trapezoid is 10x + 6y = 39

Step-by-step explanation:

* Lets explain how to solve the problem

- The slope of the line whose end points are (x1 , y1) , (x2 , y2) is

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- The mid point of the line whose end point are (x1 , y1) , (x2 , y2) is

$(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

- The standard form of the linear equation is Ax + BC = C, where

A , B , C are integers and A , B ≠ 0

- The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

- It has two properties:

# It is parallel to both bases

# Its length equals half the sum of the base lengths

* Lets solve the problem

- The trapezoid has vertices R (-1 , 5) , S (! , 8) , T (7 , -2) , U (2 , 0)

- Lets find the slope of the 4 sides two find which of them are the

parallel bases and which of them are the non-parallel bases

# The side RS

∵ $m_{RS}=\frac{8-5}{1 - (-1)}=\frac{3}{2}$

# The side ST

∵ $m_{ST}=\frac{-2-8}{7-1}=\frac{-10}{6}=\frac{-5}{3}$

# The side TU

∵ $m_{TU}=\frac{0-(-2)}{2-7}=\frac{2}{-5}=\frac{-2}{5}$

# The side UR

∵ $m_{UR}=\frac{5-0}{-1-2}=\frac{5}{-3}=\frac{-5}{3}$

∵ The slope of ST = the slop UR

∴ ST// UR

∴ The parallel bases are ST and UR

∴ The nonparallel sides are RS and TU

- Lets find the midpoint of RS and TU to find the equation of the

median of the trapezoid

∵ The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

∵ The midpoint of RS = $(\frac{-1+1}{2},\frac{5+8}{2})=(0,\frac{13}{2})$

∵ The median is parallel to both bases

∴ The slope of the median equal the slopes of the parallel bases = -5/3

∵ The form of the equation of a line is y = mx + c

∴ The equation of the median is y = -5/3 x + c

- To find c substitute x , y in the equation by the coordinates of the

midpoint of RS

∵ The mid point of Rs is (0 , 13/2)

∴ 13/2 = -5/3 (0) + c

∴ 13/2 = c

∴ The equation of the median is y = -5/3 x + 13/2

- Multiply the two sides by 6 to cancel the denominator

∴ The equation of the median is 6y = -10x + 39

- Add 10x to both sides

∴ The equation of the median is 10x + 6y = 39

* The equation of the median of the trapezoid is 10x + 6y = 39

7 0

3 years ago