A teacher wants to test the effectiveness of two

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

Solve for x in the diagram below.

marshall27 [118]

Vertical angles are congruent, so:

$4x+50=150\\\\4x=100\\\\x=\boxed{25}$

4 0

2 years ago

If f(x)=3x+5/x, what is f(a+2)? A. 3(f(a))+5/f(a)+2 B. 3a+5/a+2 C. 3(a+2)+5/a+2

natali 33 [55]

f(x) = 3x + 5/x

f(a + 2) = 3.(a + 2) + 5/(a + 2)

Alternative C.

5 0

3 years ago

Hello Brainly, Please send the details for the area of each shape. Sincerely, Leanoriaia

Ainat [17]

This type of problems are solved by partitioning the figure into rectangles, whose areas we can find easily.

Picture 1)

From the point denoted by E, draw two altitudes towards the sides AG and AC.

Since ED is 3, then BC is 3. Thus, AB=4-3=1.

Since EF is 1, then HG is 1, so AH=3-1=2.

The area of the figure is equal to the area of rectangle ACDH + the area of the square HEFG.

That is, the area of the figure is 4*2+1*1=8+1=9 (square in).

Picture 2)

As in the first part, draw the altitudes EB and EH from point E.

GF is 5, so AB is 5. This means that BC=7-5=2 (in).

FE is 3, so GH is 3, that is HA=6-3=3.

Thus, Area(figure)=Area(ABFG)+Area(BCDE)=5*6+2*3=30+6=36 (square in.)

Answer: 9 (square in); 36 (square in.)

7 0

3 years ago

Simplify five to the power of -7×5 to power of 12×5 the power of -2

Leto [7]

The answer to your question is

5 0

3 years ago