Please help???? Geometry

Mathematics

2 answers:

AlladinOne [14]3 years ago

7 0

The equation for a line is y = mx + b. The m represents slope, and the b represents the y-intercept.

Finding the slope (m): rise over run or change in y over change in x

rise 2 1
-------- = --- = ---
run 8 4

Equation: y = 1/4 x + b

Finding y intercept:

Since the line crosses the y axis at (0,0), the y intercept is 0.

Equation: y = 1/4x

Answer: E. y = 1/4x

Musya8 [376]3 years ago

6 0

Any line can be expresses as:

y=mx+b where m=slope=(y2-y1)/(x2-x1) and b=y-intercept (value of y when x=0)

First find the slope:

m=(2-0)/(8-0)=2/8=1/4 so we have thus far:

y=0.25x+b, we solve for b using any point on the line, (8,2)

2=0.25(8)+b

2=2+b

0=b

So the line is:

y=.25x which they might also express as y=x/4

The answer is E. (1/4)x

You might be interested in

Plz answer all these questions for me I really need help

Ber [7]

A is 400

E I don't know soory

4 0

3 years ago

A person draws a card from a hat. Each card is one color with the following probabilities of being drawn: 1/10 for red, 1/5 for

Kay [80]

Answer:

$\dfrac{1}{4}$

Step-by-step explanation:

We are given that a card is drawn from a hat.

Probability that a red card is drawn, P(red) = $\frac{1}{10}$

Probability that a blue card is drawn, P(blue) = $\frac{1}{5}$

Probability that a black card is drawn, P(black) = $\frac{1}{20}$

Probability that a pink card is drawn, P(pink) = $\frac{1}{5}$

All these events are mutually exclusive i.e. they do not have anything in common.

For any two events A and B which are mutually exclusive, with probability P(A) and P(B), the probability that any one of the two events occur:

$P(A\cup B) = P(A) + P(B)$

Here event A can be thought of as drawing a blue card and

event B can be thought of as drawing a black card

So, P(blue or black) = P(blue) + P(black)

$\Rightarrow \dfrac{1}{5}+\dfrac{1}{20}\\\Rightarrow \dfrac{4+1}{20}\\\Rightarrow \dfrac{5}{20}\\\Rightarrow \dfrac{1}{4}$

So, the probability that a blue or black card is drawn from the hat:

$\dfrac{1}{4}$

8 0

3 years ago

How to solve -2x^2 + 3x - 9 = 0 Can you also give me the steps I really don't understand this?

Maksim231197 [3]

The "no thinking required" way to solve it is using the quadratic formula. That formula tells you the solution to

$ax^2+bx+c=0$

is given by

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Your equation has a=-2, b=3, c=-9, so the solution is

$x=\dfrac{-3\pm \sqrt{3^2-4\cdot (-2)\cdot (-9)}}{2\cdot (-2)}\\\\=\dfrac{3\pm \sqrt{9-72}}{4}=\dfrac{3\pm 3\sqrt{-7}}{4}\\\\x=\dfrac{3}{4}\pm i\dfrac{3\sqrt{7}}{4}$