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3 years ago

What is the equation of the following line written in slope intercept form? (-5,-1)

Mathematics

1 answer:

solmaris [256]3 years ago

4 0

Answer:

$\large\boxed{y=-\dfrac{2}{3}x-\dfrac{13}{3}}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

From the graph we have two points (-5, -1) and (-2, -3).

Look at the picture.

Calculate the slope:

$m=\dfrac{-3-(-1)}{-2-(-5)}=\dfrac{-2}{3}=-\dfrac{2}{3}$

Put it to the equation in slope-intercept form:

$y=-\dfrac{2}{3}x+b$

We can't read the y-intercept from the graph. Therefore put the coordinates of the point (-5, -1) to the equation and calculate b:

$-1=-\dfrac{2}{3}(-5)+b$

$-1=\dfrac{10}{3}+b$ subtract 10/3 from both sides

$-\dfrac{3}{3}-\dfrac{10}{3}=b\to b=-\dfrac{13}{3}$

Finally:

$y=-\dfrac{2}{3}x-\dfrac{13}{3}$

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2 years ago

What is a negative 3 cubed written as a fraction

Advocard [28]

When you have an exponent that is negative, it means to take reciprocal of the fraction then take it the power.

Some examples:

5^-2 = 1/5^2 = 1/25
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8 0

3 years ago

Read 2 more answers

Find x 5[x^2]+5[x]-x^2-x=2004

a_sh-v [17]

The answer is either $x=\frac{-1}{2}+\frac{1}{2}\sqrt{2005}\\$ or $x=\frac{-1}{2}+\frac{-1}{2}\sqrt{2005}$

To solve: Let's solve your equation step-by-step.

5x2+5x−x2−x=2004

Step 1: Simplify both sides of the equation.

4x2+4x=2004

Step 2: Subtract 2004 from both sides.

4x2+4x−2004=2004−2004

4x2+4x−2004=0

Step 3: Use quadratic formula with a=4, b=4, c=-2004.

x= −b±√b2−4ac /2a

x= −4±√32080 /8

$x=\frac{-1}{2}+\frac{1}{2}\sqrt{2005}\\$ or $x=\frac{-1}{2}+\frac{-1}{2}\sqrt{2005}$

4 0

3 years ago

Read 2 more answers

Solve for r 2r = 7/10

deff fn [24]

Answer:

7/20

Step-by-step explanation:

given,

2r = 7/10

r= 7/10*1/2

r= 7/20

6 0

2 years ago

Write functions for each of the following transformations using function notation. Choose a different letter to represent each f

joja [24]

Answer:

1. Translation: g(x) = f(x-a)+b.

2. Reflection around y-axis: h(x) = f(-x)

3. Reflection around x-axis: k(x) = -f(x)

4. Rotation of 90° : $R_{90} (x,y)=(-y,x)$

5. Rotation of 180° : $R_{180} (x,y)=(-x,-y)$ .

6. Rotation of 270° : $R_{180} (x,y)=(y,-x)$ .

Step-by-step explanation:

Let us assume that the transformations namely translation, reflection are applied to a function f(x) and the rotation is applied to the point ( x,y ).

So, according to the options:

We know that 'translation moves the image in horizontal and vertical direction'.

1. As we have to translate the function f(x) 'a' units to the right and 'b' units up. So, the new form of the function becomes g(x) = f(x-a)+b.

Further, we know that 'reflection means to flip the image around a line'.

2. As, we have to reflect the function f(x) around y-axis. The new form of the function is h(x) = f(-x).

3. As, we have to reflect the function f(x) around x-axis. The new form of the function is k(x) = -f(x).

Since, 'rotation turns the image around a point to a certain degree'.

4. As, we have to rotate ( x,y ) counter-clockwise to 90° about the origin, the new form of the function is $R_{90} (x,y)=(-y,x)$ .

5. As, we have to rotate ( x,y ) counter-clockwise to 180° about the origin, the new form of the function is $R_{180} (x,y)=(-x,-y)$ .

6. As, we have to rotate ( x,y ) counter-clockwise to 270° about the origin, the new form of the function is $R_{180} (x,y)=(y,-x)$ .

5 0

3 years ago