Ask question

Ask question

All categories

3 years ago

6

In which quadrant do the points have negative x-coordinates and negative y-coordinates?

2 answers:

sashaice [31]3 years ago

7 0

Say we have (-5, -7).

You would go 5 left, and then 7 down. That would land you in Quadrant III.
The top right quadrant is quadrant I, the top left is quadrant II, the bottom left is Q. III, the bottom right is Q. IV.

posledela3 years ago

5 0

The answer is Quadrant III.

You might be interested in

A cube has a surface area of 864 square inches.

slamgirl [31]

The answer is (D) or 1728 In³

6 0

3 years ago

Read 2 more answers

You scored a 95% on your math quiz.The quiz was out of 60 points.How many points did you get

Maurinko [17]

Yo scored 57 out of 60

4 0

3 years ago

Hi, I really need help with my homework please help

8090 [49]

Number one is 68 you have to divide on this download photomath its better because all you have to do is take a picture and it shows you how to do it ;)

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=2x-y%3D4%5C%5Cy-x%3D4" id="TexFormula1" title="2x-y=4\\y-x=4" alt="2x-y=4\\y-x=4" align="absmi

Flura [38]

Here is the answer. Hope this helps!

5 0

1 year ago

1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera

Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

$SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}$

$SP \approx 8.246$

$PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}$

$PA \approx 8.246$

$AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}$

$AZ \approx 8.246$

$ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}$

$ZS \approx 8.246$

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

$m_{SA} = \frac{6-(-4)}{6-(-4)}$

$m_{SA} = 1$

$m_{PZ} = \frac{4-(-2)}{-2-4}$

$m_{PZ} = -1$

Since $m_{SA}\cdot m_{PZ} = -1$ , diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

$M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)$

$M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)$

$M_{SA} = (-2,-2)+(3,3)$

$M_{SA} = (1,1)$

$M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)$

$M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)$

$M_{PZ} = (2,-1)+(-1,2)$

$M_{PZ} = (1,1)$

Then, the diagonals SA and PZ bisect each other.

8 0

2 years ago