What is the slope of a line segment PQ with coordinates P(-2.5) and Q(-1,1)? show your work.

Mathematics

2 answers:

PolarNik [594]3 years ago

8 0

Answer:

-4

Step-by-step explanation:

<u>Given</u>

Line segment PQ with points P(-2.5) and Q(-1,1)
Slope = ?

<u>Slope formula</u>

m = (y₂ - y₁)/(x₂ - x₁)
m = (1 - 5)/(-1 - (-2)) = -4/1 = -4
Slope is -4

lisov135 [29]3 years ago

4 0

We can use the given points to solve.

Slope formula: y2-y1/x2-x1

(1-5)/-1-(-2))

-4/1

-4

Best of Luck!

You might be interested in

Please help me thank you I will give you brainly

aliya0001 [1]

The 3rd selection, "Given" is appropriate.

4 0

3 years ago

Plssss help quick i need it for sum important

Vanyuwa [196]

(a) The product of $(2x-4)\ and\ (3x^2-x+4)$ is calculated to be $=6x^3-14x^2+12x-16$

(b)The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

<u>Step-by-step explanation:</u>

$(2x-4)(3x^2-x+4)\\\\=(2x.3x^2)+(2x.-x)+(2x.4)+(-4.3x^2)+(-4.-x)+(-4.4)\\\\=6x^3-2x^2+8x-12x^2+4x-16\\\\=6x^3-14x^2+12x-16$

$(4x-2)(3x^2-x+4)\\\\=(4x.3x^2)+(4x.-x)+(4x.4)+(-2.3x^2)+(-2.-x)+(-2.4)\\\\=12x^3-4x^2+16x-6x^2+2x-8\\\\=12x^3-10x^2+18x-8$

The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

7 0

3 years ago

1. Factor and simplify: cos² xcsc² x-cos²xcot² x

Vera_Pavlovna [14]

Answer:

Cos² x

Step-by-step explanation:

<h3>Trigonometry:</h3>

$\sf Cos^2 \ x *Csc^2 \ x-Cos^2 \ x *Cot^2 \ x = Cos^2 \ x (Csc^2 \ x - Cot^2 \ x)$

$\sf = Cos^2 \ x \left(\dfrac{1}{Sin^2 \ x} - \dfrac{Cos^2 \ x}{Sin^2 \ x}\right)\\\\= Cos^2 \ x \left(\dfrac{1-Cos^2 \ x}{Sin^2 \ x}\right)\\\\\boxed{\bf Indentity: \ 1 - Cos^2 \ x = Sin^2 \ x}\\\\\\= Cos^2 \ x \left(\dfrac{Sin^2 \ x}{Sin^2 \ x}\right)\\\\= Cos^2 \ x$