3 years ago

Given line segment AB, the coordinates of A are (3,-5) and the coordinates midpoint are (5,-1), what are the coordinates for B?

Mathematics

1 answer:

dsp733 years ago

6 0

Answer:

Step-by-step explanation:

Point P(5, -1) being the midpoint of AB means that $x_P-x_A=x_B-x_P$

$5-3=x_B-5\\\\2+5=x_B\\\\x_B=7$

and $y_P-y_A=y_B-y_P$

$-1-(-5)=y_B-(-1)\\\\-1+5=y_B+1\\\\y_B=3$

You might be interested in

Obtain the general solution to the equation. (x^2+10) + xy = 4x=0 The general solution is y(x) = ignoring lost solutions, if any

alukav5142 [94]

Answer:

$y(x)=4+\frac{C}{\sqrt{x^2+10}}$

Step-by-step explanation:

We are given that a differential equation

$(x^2+10)y'+xy-4x=0$

We have to find the general solution of given differential equation

$y'+\frac{x}{x^2+10}y-\frac{4x}{x^2+10}=0$

$y'+\frac{x}{x^2+10}y=4\frac{x}{x^2+10}$

Compare with

$y'+P(x) y=Q(x)$

We get

$P(x)=\frac{x}{x^2+10}$

$Q(x)=\frac{4x}{x^2+10}$

I.F= $e^{\int\frac{x}{x^2+10} dx}=e^{\frac{1}{2}ln(x^2+10)}$

$e^{ln\sqrt(x^2+10)}=\sqrt{x^2+10}$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{x^2+10}\times \sqrt{x^2+10} dx+C$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{\sqrt{x^2+10}}+C$

$y\cdot \sqrt{x^2+10}=4\sqrt{x^2+10}+C$

$y(x)=4+\frac{C}{\sqrt{x^2+10}}$

6 0

3 years ago

at an aquarium six out of 18 deliveries are plants out of 15 delivery's in one week how manybare plants

earnstyle [38]

Do six multiply fifteen which equals to 90

5 0

3 years ago

Solve the system of equations.

Helga [31]

The answer is x=1 y=-1 and z=1. If you plug in the variables you get
4-3+5 and that equals 6. So that’s the answer

7 0

2 years ago

I need help plotting before 7:00.

Lemur [1.5K]

J: (4,0)

M:(-1,1)

E:(4,-1)

X:(1,0)

5 0

3 years ago

Can anyone do 25-30 (do not understand)

Olenka [21]

Me either hope you get help

7 0

3 years ago