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3 years ago

15

What is the slope of the line on the graph?

2 answers:

Stella [2.4K]3 years ago

8 0

Answer:

none.

Step-by-step explanation:

if you examine the dots and solve, they are not equivalent

also it doesn't pass through the orgin

dexar [7]3 years ago

5 0

Given points (1,5) and (0,-1)
(-1-5)/(0-1) = -6/-1 = 6
The slope is 6

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Solve the equation. 33 = p – 6.71<br> A. –39.71 <br> B. –26.29 <br> C. 39.71 <br> D. 26.29

jeka57 [31]

<u>Answer</u>

C. 39.71

<u>Explanation</u>

33 = p - 6.71

The first step is to make the like terms to be on the same side.

Add 6.71 on both sides of the eqution

33 + 6.71 = p - 6.71 + 6.71

39.71 = p

∴ p = 39.71

4 0

3 years ago

In the box, type the number that will correctly complete the sentence.

Tpy6a [65]

Answer:

1

Step-by-step explanation:

One. For example, if we chose 5 as the number, then the multiplicative inverse would be 1/5. Multiplying 5 by 1/5 results in one (1).

7 0

3 years ago

Gnom [1K]

$63.2 80% of 79 = 63.2

5 0

3 years ago

Graph the solution of the inequality: 3/7 (35x-14) ≤ 21x/2 + 3

hammer [34]

Hello there!

In order to graph the solution, we need to solve it first.

So let's start....

3/7(35x - 14) ≤ 21x/2 + 3

15x -6 ≤ 21x/2 + 3

15x - 21x/2 ≤ 3 + 6

9x/2 ≤ 9

x ≤ 9* 2/9

x ≤ 18/9

x ≤ 2

See the picture for the graph

Please let me know if you have questions about the answer.

Thanks!

5 0

3 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago