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2 years ago

How do i do rise over run if the slope is not a fraction

2 answers:

7 0

Ok for example let’s use 2 as the slope. You are going to put the 2 over 1 so it will look like this 2/1. And that would be your slope!

Sloan [31]2 years ago

3 0

So let’s say the slope is 5 then you would just put a 1 under it like this .... 5/1

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Write a slope- intercept equation for a line with given characters m=-4/5, passes through (-4,-5)

aleksandr82 [10.1K]

$\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-5})~\hspace{10em} \stackrel{slope}{m}\implies -\cfrac{4}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-5)}=\stackrel{m}{-\cfrac{4}{5}}[x-\stackrel{x_1}{(-4)}]\implies y+5=-\cfrac{4}{5}(x+4) \\\\\\ y+5=-\cfrac{4}{5}x-\cfrac{16}{5}\implies y=-\cfrac{4}{5}x-\cfrac{16}{5}-5\implies y=-\cfrac{4}{5}x-\cfrac{41}{5}$

6 0

2 years ago

For the equation ae^ct=d, solve for the variable t in terms of a,c, and d. Express your answer in terms of the natural logarithm

saveliy_v [14]

We have been given an equation $ae^{ct}=d$ . We are asked to solve the equation for t.

First of all, we will divide both sides of equation by a.

$\frac{ae^{ct}}{a}=\frac{d}{a}$

$e^{ct}=\frac{d}{a}$

Now we will take natural log on both sides.

$\text{ln}(e^{ct})=\text{ln}(\frac{d}{a})$

Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$ct\cdot \text{ln}(e)=\text{ln}(\frac{d}{a})$

We know that $\text{ln}(e)=1$ , so we will get:

$ct\cdot 1=\text{ln}(\frac{d}{a})$

$ct=\text{ln}(\frac{d}{a})$

Now we will divide both sides by c as:

$\frac{ct}{c}=\frac{\text{ln}(\frac{d}{a})}{c}$

$t=\frac{\text{ln}(\frac{d}{a})}{c}$

Therefore, our solution would be $t=\frac{\text{ln}(\frac{d}{a})}{c}$ .

5 0

3 years ago

Solve the following system of equations: 2x + 4y − 3z = −7 3x + y + 4z = −12 x + 3y + 4z = 4

lukranit [14]

2x + 4y − 3z = −7 ---> first equation

3x + y + 4z = −12 ---> second equation

x + 3y + 4z = 4 ---> third equation

We try to eliminate z and make two equations

Multiply first equation by 4 and second equation by 3. then add both equations

8x + 16y -12z = -28

9x + 3y +12z = -36

----------------------------

17x + 19y = -64 ---------------> equation 4

multiply third equation by -1 and add second and third equation

3x + y + 4z = −12

-x - 3y - 4z = -4

-------------------------------------

2x -2y = -16 ---------------> equation 5

Multiply equation 4 by 2 and equation by 19

34x + 38y = -128

38x - 38y = -304

------------------------------------

72x = -432

Divide by 72 on both sides

x = -6

use equation 2x -2y = -16 ---------------> equation 5

2(-6) - 2y = -16

-12 -2y = -16

-2y = -4

So y = 2

Now plug in 2 for y and -6 for x

Using any one of the three equations

x + 3y + 4z = 4

-6 + 3(2) +4z = 4

4z = 4 (divide by 4 on both sides)

So z=1

So x= -6, y = 2 and z= 1

4 0

3 years ago

Read 2 more answers

What is the domain of y=log5x

aliina [53]

The answer is both all real numbers less than 0 and all real numbers greater than 0

3 0

3 years ago

Read 2 more answers

13. Which system of inequalities has no solution?

Anit [1.1K]

Answer:

Option A

Step-by-step explanation:

We have y < 2x - 4 and y > 2x + 1.

=> 2x - 4 > 2x + 1

=> -4 > 1

Since this inequality is false for all real values of x, there is no solution for the system of inequalities.

8 0

3 years ago