Look at the picture please.

The coordinates of point A are (p, q) and coordinates of point B are (p+2q, q+2p). Provide your complete solutions and proofs in

Dafna1 [17]

Answer:

The mid-point (p + q , q + p) of AB is the same distance from the x-axis and the y-axis

Step-by-step explanation:

* Lets explain how to solve the problem

- Any point will be equidistant from the x-axis and the y-axis must have

equal coordinates

- Ex: point (4 , 4) is the same distance from the x-axis and the y-axis

because the distance from the x-axis to the point is 4 (y-coordinate)

and the distance from the y-axis and the point is 4 (x-coordinate)

- If (x , y) is the mid-point of a segment its endpoints are

$(x_{1},y_{1})$ and $(x_{2},y_{2})$ , then

$x=\frac{x_{1}+x_{2}}{2}$ and $y=\frac{y_{1}+y_{2}}{2}$

* Lets solve the problem

∵ Point A has coordinates (p , q)

∵ Point B has coordinates (p + 2q , q + 2p)

- The mid-point of AB is (x , y)

∵ $x=\frac{p+p+2q}{2}$

∴ $x=\frac{2p+2q}{2}$

- Take 2 as a common factor from the terms of the numerator

∴ $x=\frac{2(p+q)}{2}$

- Divide up and down by 2

∴ x = p + q

∵ $y=\frac{q+q+2p}{2}$

∴ $y=\frac{2q+2p}{2}$

- Take 2 as a common factor from the terms of the numerator

∴ $y=\frac{2(q+p)}{2}$

- Divide up and down by 2

∴ y = q + p

∴ The mid point of AB is (p + q , q + p)

- p + q is the same with q + p

∵ The x-coordinate of the mid point of AB is p + q

∵ The y-coordinate of the mid point of AB is q + p

∵ p + q = q + p

∴ The coordinates of the mid-point of AB are equal

- According the explanation above

∴ The mid-point (p + q , q + p) of AB is the same distance from the

x-axis and the y-axis

8 0

3 years ago

Matthew bought snacks for soccer team. He bought a bag of apple slices for $5.65, and he bought a 12 pack of Gatorade bottles. T

agasfer [191]

Answer:

6.48/12=B

Step-by-step explanation:

B=6.48

6.48/12=0.54

Each Gatorade bottle costs 54 cents.

5 0

3 years ago

Read 2 more answers

Subtraction (must show work) 1. d - 27 = 45 2. h - 114 = 28

Arturiano [62]

Answer:

Step-by-step explanation:

1.

d - 27 = 45

add 27

d = 72

2.

h - 114 = 28

add 114

h = 142

3.

-4 + x = 15

add 4

x = 19

4.

-39 + g = 72

add 39

g = 111

7 0

2 years ago

Find gradient xe^y + 4 ln y = x² at (1, 1)

cricket20 [7]

$xe^y+4\ln y=x^2$

Differentiate both sides with respect to x, assuming y = y(x).

$\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}$

$\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x$

$\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

$e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

Solve for dy/dx :

$e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x$

$\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}$

If y ≠ 0, we can write

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}$

At the point (1, 1), the derivative is

$\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}$

4 0

3 years ago

A kite flying in the air has a 12 line attached to it. Its line is pulled taut and casts a 9 shadow. Find the height of the kite

alexira [117]

Answer:

8 (7.94)

Step-by-step explanation:

You can think of it as a geometry problem.

What is formed here is a triangle, which sides ate: the line, the line's shadow, and the height from the ground to the kite (here I attach a drawing).

What you need to find is the height. We will call it H.

As the triangle formed is a right one, we can use Pitágoras' theorem. The height H squared plus the squared of the shadow is equal to the squared of the line (the hypotenuse). This is:

H^2 + 9^2 = 12^2

H^2 + 81= 144

H^2 = 63

Applying squared root in both sides

H = √63

H = 7,94

So, the height is approximately 8.

4 0

3 years ago