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3 years ago

What is slope and how is it determined on a graph

1 answer:

3 0

A slope is the change in position with respect to x and y. Use y2-y1 over x2-x1. The 2s and 1s are supscripts in this equation.

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Multiply by 4, then divide by 2 First term: 60

harina [27]

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3 0

3 years ago

LekaFEV [45]

Answer: $y-1=\dfrac32(x+3)$

Step-by-step explanation:

Slope of a line passes through (a,b) and (c,d) = $\dfrac{d-b}{c-a}$

In graph(below) given line is passing through (-2,-4) and (2,2) .

Slope of the given line passing through (-2,-4) and (2,2) = $\dfrac{-4-2}{-2-2}=\dfrac{-6}{-4}=\dfrac{3}{2}$

Since parallel lines have equal slope . That means slope of the required line would be .

Equation of a line passing through (a,b) and has slope m is given by :_

(y-b)=m(x-a)

Then, Equation of a line passing through(-3, 1) and has slope = is given by

$(y-1)=\dfrac32(x-(-3))\\\\\Rightarrow\ y-1=\dfrac32(x+3)$

Required equation: $y-1=\dfrac32(x+3)$

8 0

3 years ago

Please help me with this , it would mean alot

algol [13]

Answer:

Step-by-step explanation:

21-26. a = v - u / t

=> a = 47 - 19 / 5

=> a = 28 / 5

=> a = 5.6 m/s²

27-32. avg. speed = distance / time

=> 40 m / 10 s

=> 4 m/s

33-38. velocity = distance / time

=> v = 150 / 40

=> v = 15 / 4

=> v = 3.75 m/s

4 0

3 years ago

Solve for g.<br> g/6 - -53 = 57

Mashcka [7]

Step-by-step explanation:

g/6 - -53 = 57

g/6 + 53 = 57

g/6 = 4

g = 4×6 = 24

5 0

2 years ago

I need help asap pls and thank you ;)

olga55 [171]

Answer:

$\text{Length of AB is }\frac{ah}{a+h}$

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP (∵ corresponding angles)

∠NDC=∠NPM (∵ corresponding angles)

By AA similarity rule, ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

$\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\$