All categories

2 years ago

Write the equation in slope-intercept form: x-4y=8

Mathematics

2 answers:

ad-work [718]2 years ago

4 0

Answer:

$y = \frac{1}{4} x - 2$

Step-by-step explanation:

First thing to do is siolate the y variable. We do this by subtracting x from both sides of the equation in this particular equation.

$x - 4y = 8 \\ - x \: \: \: - x \\ - 4y = - x + 8$

Now we change the -4y into just y using division. We have to divide boths sides by -4 to do so.

$- 4y = - x + 8 \\ \div ( - 4) \: \: \div ( - 4) \\ y = \frac{1}{4}x - 2$

There it is in slope intercept form.

alisha [4.7K]2 years ago

3 0

Answer:

y=1/4x-2

Step-by-step explanation:

slope intercept form is y=mx+b

your equation right now is in standard form: ax+by=c

equation: x-4y=8

what we want to do is isolate y on one side

subtract x from both sides

-4y=-x+8

divide by 4 on both sides

y=1/4x-2

Hope this helps! :)

You might be interested in

Arthur's dog, Shep, had to go on a diet. After three months of dieting, Shep had lost 16% of his weight, which was 24 pounds. Ho

lidiya [134]

Answer: 150 pounds

Step-by-step explanation:

Let her initial weight be represented by x.

Since Shep had lost 16% of his weight, which was 24 pounds. Shep weight before he was put on a diet would be:

16% × x = 24

0.16 × x = 24

0.16x = 24

x = 24/0.16

x = 150 pounds

5 0

2 years ago

-5 -4 -3 -2 -1 0 1 2 3 4 5 -1 <3

Gelneren [198K]

Answer:

-1

Step-by-step explanation:

.........................................................................................

4 0

2 years ago

Laura's job is 4 1/2 miles away. If she starts walking to her job at 9:00 a.m. and is scheduled to start working at 11:00 a.m.,

irina [24]

Answer:

Laura has to walk at 2 1/4 miles per hour or faster to arrive to her job on time.

Step-by-step explanation:

1. Let's review the data given to us for solving the question:

Distance from Laura's home to her job = 4 1/2 miles

Time when Laura starts to walk to her job : 9:00 a.m.

Time when Laura is scheduled to start working : 11:00 a.m.

2. Will she arrive at her job on time?

Time Laura has to complete the distance from her home to her job = 2 hours (11 -9)

Speed that Laura need to walk to arrive to her job on time = Distance from Laura's home to her job/Time Laura has to complete the distance from her home to her job

Speed that Laura need to walk to arrive to her job on time = 4 1/2 miles/2 hours = (4 1/2)/2 = (9/2)/2 = 9/2 * 1/2 = 9/4 = 2 1/4

Speed that Laura need to walk to arrive to her job on time = 2 1/4 miles per hour.

Laura has to walk at 2 1/4 miles per hour or faster to arrive to her job on time.

7 0

2 years ago

All I need is the answer Please and thank you

Alex73 [517]

Should be 40ft. If not, I dearly apologize.

5 0

2 years ago

Read 2 more answers

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

kozerog [31]

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$ .

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$ , $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$ . However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$ . For any two elements $a,\, b \in S$ , $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$ .

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

Reflexivity: for any $a \in S$ , the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$ .)

Symmetry: for any $a,\, b \in S$ , $a \sim b$ if and only if $b \sim a$ . In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$ , or neither is in $R\!$ .

Transitivity: for any $a,\, b,\, c \in S$ , if $a \sim b$ and $b \sim c$ , then $a \sim c$ . In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$ , then $(a,\, c)$ also needs to be in $R\!$ .

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$ ) in this question is indeed reflexive. $(1,\, 1)$ , $(2,\, 2)$ , and $(3,\, 3)$ (one pair for each element of $S$ ) are all elements of $R\!$ .

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$ .

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$ . However, $(3,\, 2) \not \in R$ . In other words, under relation $R\!$ , $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$ .

3 0

2 years ago