All categories

3 years ago

Write the following relation as a linear equation in standard form. Include a space between terms and operations.

Mathematics

1 answer:

kow [346]3 years ago

4 0

Answer:

In the form of

Y= mx+c

Y= 1/2x +2

m = 1/2

Step-by-step explanation:

A linear equation in it's standard form is in the format

Y= mx+c

Where m is the slope and c is the y intercept

Let's use these two points to determine both the slope and the equation

(2, 3), (4,4)

Slope= (y2-y1)/(x2-x1)

Slope= (4-3)/(4-2)

Slope= 1/2

Equation of the linear function

(Y-y1)/(x-x1)= m

(Y-3)/(x-2)= 1/2

2(y-3) = x-2

2y -6 = x-2

2y= x-2+6

2y= x+4

Y= 1/2x +2

You might be interested in

Please help with this

Ghella [55]

5/1 because I know it

7 0

3 years ago

find the values of the six trigonometric functions for angle theta in standard position if a point with the coordinates (1, -8)

frutty [35]

Answer:

cosФ = $\frac{1}{\sqrt{65}}$ , sinФ = $-\frac{8}{\sqrt{65}}$ , tanФ = -8, secФ = $\sqrt{65}$ , cscФ = $-\frac{\sqrt{65}}{8}$ , cotФ = $-\frac{1}{8}$

Step-by-step explanation:

If a point (x, y) lies on the terminal side of angle Ф in standard position, then the six trigonometry functions are:

cosФ = $\frac{x}{r}$
sinФ = $\frac{y}{r}$
tanФ = $\frac{y}{x}$
secФ = $\frac{r}{x}$
cscФ = $\frac{r}{y}$
cotФ = $\frac{x}{y}$

Where r = $\sqrt{x^{2}+y^{2} }$ (the length of the terminal side from the origin to point (x, y)

You should find the quadrant of (x, y) to adjust the sign of each function

∵ Point (1, -8) lies on the terminal side of angle Ф in standard position

∵ x is positive and y is negative

→ That means the point lies on the 4th quadrant

∴ Angle Ф is on the 4th quadrant

∵ In the 4th quadrant cosФ and secФ only have positive values

∴ sinФ, secФ, tanФ, and cotФ have negative values

→ let us find r

∵ r = $\sqrt{x^{2}+y^{2} }$

∵ x = 1 and y = -8

∴ r = $\sqrt{x} \sqrt{(1)^{2}+(-8)^{2}}=\sqrt{1+64}=\sqrt{65}$

→ Use the rules above to find the six trigonometric functions of Ф

∵ cosФ = $\frac{x}{r}$

∴ cosФ = $\frac{1}{\sqrt{65}}$

∵ sinФ = $\frac{y}{r}$

∴ sinФ = $-\frac{8}{\sqrt{65}}$

∵ tanФ = $\frac{y}{x}$

∴ tanФ = $-\frac{8}{1}$ = -8

∵ secФ = $\frac{r}{x}$

∴ secФ = $\frac{\sqrt{65}}{1}$ = $\sqrt{65}$

∵ cscФ = $\frac{r}{y}$

∴ cscФ = $-\frac{\sqrt{65}}{8}$

∵ cotФ = $\frac{x}{y}$

∴ cotФ = $-\frac{1}{8}$

8 0

2 years ago

141.1933-67.53 =<br> (Divided, not minus)

Sphinxa [80]

Answer: I think its 2.09

Step-by-step explanation:

But i'm not sure.

5 0

3 years ago

Please help<br><br>Point____is plotted at -4/5

musickatia [10]

Answer:

Step-by-step explanation:

-4/5 is -0.8, which is at point C

plz brainliestt

8 0

3 years ago

Read 2 more answers

Put these points into an equation IN STANDARD FORM!!!!!!<br><br>(0,2) and (8,0)

zubka84 [21]

The equation of the line in standard form is x + 4y = 8

<h3>How to determine the line equation?</h3>

From the question, the points are given as

(0, 2) and (8, 0)

To start with, we must calculate the slope of the line

This is calculated using

Slope = (y₂ - y₁)/(x₂ - x₁)

Where

(x, y) = (0, 2) and (8, 0)

Substitute the known parameters in Slope = (y₂ - y₁)/(x₂ - x₁)

So, we have

Slope = (0 - 2)/(8 - 0)

Evaluate

Slope = -1/4

The equation of the line can be calculated using as

y - y₁ = m(x + x₁)

Where

(x₁, y₁) = (0, 2)

and

m = slope = -1/4

Substitute the known values in the above equation

So, we have the following equation

y - 2 = -1/4(x - 0)

This gives

y - 2 = -1/4x

Rewrite as

1/4x + y = 2

Multiply by 4

x + 4y = 8

Hence, the line has an equation of x + 4y = 8