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3 years ago

What is the slope of a line perpendicular to the line whose equation is 2x+4y=-642x+4y=−64. Fully simplify your answer.

Mathematics

1 answer:

labwork [276]3 years ago

5 0

9514 1404 393

Answer:

Step-by-step explanation:

Solving the given equation for y, you have ...

2x +4y = -64

4y = -2x -64

y = -1/2x -16

The coefficient of x is the slope of the given line: -1/2. The slope of the perpendicular line is the opposite reciprocal of this:

-1/(-1/2) = 2

The slope of the perpendicular line is 2.

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How do uou do this problem

Ronch [10]

The first step to solving an equation like this is to find the slope of a line that will be perpendicular to the line given. The slope of a line that's perpendicular to another line is the negative reciprocal. The negative reciprocal of -1/5 is 5. So, so far our equation is y = 5x + b. Now, to find what b is equal to, we should substitute the values of x and y from the point (1,2) since we know that our line goes through the point. Our equation becomes:

2 = 5 + b
b= -3

That means that the equation of our new line is y = 5x - 3

4 0

3 years ago

Graph the system & write its solution <br> 2x+y=-4<br> Y=-1/2x-1

xz_007 [3.2K]

Answer:

The solution of system of equation is (-2,0)

Step-by-step explanation:

Given system of equation are

Equation 1 : 2x+y=(-4)

Equation 2 : y+ $\frac{1}{2}$ x=(-1)

To plot the equation of line, we need at least two points

For Equation 1 : 2x+y=(-4)

Let x=0

2x+y=(-4)

2(0)+y=(-4)

y=(-4)

Let x=1

2x+y=(-4)

2(1)+y=(-4)

y=(-6)

Therefore,

The required points for equation is (0,-4) and (1,-6)

For Equation 2 : y+ $\frac{1}{2}$ x=(-1)

Let x=0

y+ $\frac{1}{2}$ x=(-1)

y+ $\frac{1}{2}$ (0)=(-1)

y=(-1)

Let x=2

y+ $\frac{1}{2}$ x=(-1)

y+ $\frac{1}{2}$ (2)=(-1)

y=(-2)

The required points for equation is (0,-1) and (2,-2)

Now, plot the graph using this points

From the graph,

The red line is equation 1 and blue line is equation 2

Since. The point of intersection is solution of system of equations

The solution of system of equation is (-2,0)

6 0

3 years ago

Analyze the graph of the cube root function shown

Greeley [361]

The values of a, h, and k in the general equation is h=1, k=4 and a=2

<h3>What is parent function?</h3>

The function which can be transformed into different ways is the parent function.

On a coordinate plane, a cubic function approaches the x-axis in quadrant 2, crosses the y-axis at (0, 2), has an inflection point at (1, 4), and the goes through (2, 6).

The given function is

y =a³ √(x -h) + k

Put the coordinate (0,2) in the equation, we get

2 = a³ √(0 -h) + k..................(1)

Plug (1, 4), we have

4 = a³ √(1 -h) + k..................(2)

Plug (2, 6), we have

6 = a³ √(2 -h) + k..................(3)

On solving these three equations, we have

h=1

k=4

a=2

Thus, value of h is 1, k is 4 and a is 2.

Learn more about parent function

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7 0

1 year ago

What is the area of a gameboard if it is 9 1/2 inches by 11 3/4 inches

Reil [10]

You just need to multiply 9 1/2 by 11 3/4. simplify these into decimals: 9.5 and 11.75

Then do 9.5*11.75=111.625

4 0

3 years ago

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

3 years ago