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3 years ago

You start at (2, 5). You move left 5 units and right 8 units. Where do you end?

Mathematics

2 answers:

lianna [129]3 years ago

7 0

Answer:

(5,5)

Step-by-step explanation:

left 5 is -5 and right 8 is +8.

-5+8=3

2+3=5

When talking about left and right, you change the x-coordinate, when up and down, y-coordinate.

Overall, (5,5)

Vitek1552 [10]3 years ago

4 0

Answer:

5,5

Step-by-step explanation:

since we are not changing the height of the point and just moving side to side, the y intercept does not change. Once you move 5 units to the left, your new point is -3,5 but then when you move 8 to the right it switches to positive again and the new point is 5,5

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3 years ago

Read 2 more answers

∆ABC has vertices A(–2, 0), B(0, 8), and C(4, 2)

Natali [406]

Answer:

Part 1) The equation of the perpendicular bisector side AB is $y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) The equation of the perpendicular bisector side BC is $y=\frac{2}{3}x+\frac{11}{3}$

Part 3) The equation of the perpendicular bisector side AC is $y=-3x+4$

Part 4) The coordinates of the point P(0.091,3.727)

Step-by-step explanation:

Part 1) Find the equation of the perpendicular bisector side AB

we have

A(–2, 0), B(0, 8)

step 1

Find the slope AB

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{8-0}{0+2}$

$m=4$

step 2

Find the slope of the perpendicular line to side AB

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-\frac{1}{4}$

step 3

Find the midpoint AB

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+0}{2},\frac{0+8}{2})$

$M(-1,4)$

step 4

Find the equation of the perpendicular bisectors of AB

the slope is $m=-\frac{1}{4}$

passes through the point $(-1,4)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$4=(-\frac{1}{4})(-1)+b$

solve for b

$b=4-\frac{1}{4}$

$b=\frac{15}{4}$

$y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) Find the equation of the perpendicular bisector side BC

we have

B(0, 8) and C(4, 2)

step 1

Find the slope BC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-8}{4-0}$

$m=-\frac{3}{2}$

step 2

Find the slope of the perpendicular line to side BC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=\frac{2}{3}$

step 3

Find the midpoint BC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{0+4}{2},\frac{8+2}{2})$

$M(2,5)$

step 4

Find the equation of the perpendicular bisectors of BC

the slope is $m=\frac{2}{3}$

passes through the point $(2,5)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$5=(\frac{2}{3})(2)+b$

solve for b

$b=5-\frac{4}{3}$

$b=\frac{11}{3}$

$y=\frac{2}{3}x+\frac{11}{3}$

Part 3) Find the equation of the perpendicular bisector side AC

we have

A(–2, 0) and C(4, 2)

step 1

Find the slope AC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-0}{4+2}$

$m=\frac{1}{3}$

step 2

Find the slope of the perpendicular line to side AC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-3$

step 3

Find the midpoint AC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+4}{2},\frac{0+2}{2})$

$M(1,1)$

step 4

Find the equation of the perpendicular bisectors of AC

the slope is $m=-3$

passes through the point $(1,1)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$1=(-3)(1)+b$

solve for b

$b=1+3$

$b=4$

$y=-3x+4$

Part 4) Find the coordinates of the point of concurrency of the perpendicular bisectors (P)

we know that

The point of concurrency of the perpendicular bisectors is called the circumcenter.

Solve by graphing

using a graphing tool

the point of concurrency of the perpendicular bisectors is P(0.091,3.727)

see the attached figure

5 0

3 years ago

The area of a rhombus whose diagonals are of lengths 10 cm and 8.2 cm is ____

ololo11 [35]

Answer:

area= 41 cm^2

Step-by-step explanation:

area=5 x 8.2 = 41 cm^2

divide the rhombus in half either way and you have two identical triangles.

area of triangle =0.5 x base x height

double that for the two triangles and you have the area of the rhombus.

8 0

2 years ago

Regular hexagon ABCDEF has vertices at A(4, 4!3), B(8, 4!3), C(10, 2!3), D(8, 0), E(4, 0) and F(2, 2!3).

marissa [1.9K]

Since the given hexagon is a regular hexagon all it's sides will be of equal length. Now, we know that the Area of any regular hexagon is given by:

$A=\frac{3\sqrt{3}}{2} a^2$

Where $A$ is the area of the regular hexagon

$a$ is the side length of the regular hexagon

Also, it's Perimeter is given by:

$P=6a$

Thus, all that we need to do is to find the side length of any one of the sides and to do that let us have a look at at the data of vertices points given and find out which points are definitely adjacent to each other and are also easy to calculate.

A quick search will yield that D(8, 0) and E(4, 0) are definitely adjacent to each other.

Please check the attached file here for a better understanding of the diagram of the original regular hexagon. Points D and E indeed are adjacent to each other.

Let us now find the distance between the points D and E using the distance formula which is as:

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

Where $d$ is the distance.

$(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of points D and E respectively. (please note that interchanging the values of the coordinates will not alter the distance $d$ )