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3 years ago

Which is the other endpoint of a line segment with one endpoint at (1, 3) and midpoint at (3,5)?

Mathematics

1 answer:

Reil [10]3 years ago

3 0

Answer:

C. (5,7)

Step-by-step explanation:

One endpoint at 1,3 and the midpoint at 3,5Your answer would be to as two to each the x and y of the midpoint. So the answer would be 5,7.

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Pls write answer as soon as possible

Tomtit [17]

Answer:

Step-by-step explanation:

11.-3(2x-8)=-12

-6x+24=-12

-6x=-12-24

x=-36/-6

x=6

12.4(6+2x)=0

24+8x=0

8x=0-24

x=-24/8

x=-3

13.3x+2x+6=-15

5x+6=-15

5x=-15-6

x=-21/5

14.4=-2(x+3)

4=-2x-6

4+6=-2x

10/-2=x

-5=x

15.27=46+2x-x

27=46+x

27-46=x

-19=x

7 0

3 years ago

Read 2 more answers

6x-2=7x+9 find x plz help

steposvetlana [31]

Answer:

$x=-11$

Step-by-step explanation:

$6x-2=7x+9$

Add 2 to both sides:

$6x-2+2=7x+9+2$

$6x=7x+11$

Subtract 7x from both sides:

$6x-7x=7x+11-7x$

$-x=11$

Divide both sides by -1:

$\frac{-x}{-1}=\frac{11}{-1}$

$x=-11$

7 0

3 years ago

Read 2 more answers

The vertices of a triangle are P(8, -5), Q(-5, -8), and R(-4, -3). What are the vertices of the

Kobotan [32]

Answer:

Step-by-step explanation:

5 0

2 years ago

A rectangular garden is 5 ft longer than it is wide. Its area is 300 ft2. What are its dimensions?

topjm [15]

Answer:

width=x

length=5x

area= 300ft2

area=lxb

300=5x(x)

300=5x^2

x^2=60

x=7.74

7.74(5)=38.7

dimensions are 38.7ft by 7.74 ft

Please mark as brainliest

5 0

2 years ago

Read 2 more answers

The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter

Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

$(18,12)$ -- False

$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

$y = 1\frac{1}{2}x$

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

The point $(1\frac{1}{2},1)$ won't be on the graph because the corresponding value of y for $x = 1\frac{1}{2}$ is $y = 2\frac{1}{4}$

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

The point $(4,6)$ would be on the graph because the corresponding value of y for $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

The point $(18,12)$ wouldn't be on the graph because the corresponding value of y for $x = 18$ is $y = 12$

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

The point $(0,0)$ would be on the graph because the corresponding value of y for $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

$y = \frac{3}{2} * \frac{5}{2}$

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

$y = 3\frac{3}{4}$

The point $(2\frac{1}{2},3\frac{3}{4})$ would be on the graph because the corresponding value of y for $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

3 0

3 years ago