What is the slope of the line that passes through the points (-1, 2) and (-4, 3)?

Mathematics

1 answer:

I am Lyosha [343]2 years ago

6 0

Answer:

-0.5

Step-by-step explanation:

(3-2)=1

(-4-1)=-5

1÷(-5)= -0.5

You might be interested in

Is this expression equivalent to 1.5?<br><br> 15.0 / 10<br> 15.0 (divided by) 10

Dmitriy789 [7]

Answer:

Yes it is true

Step-by-step explanation:

15/10 equals 1.5

4 0

2 years ago

Solve for the unknown by using the additive inverse. Type the FULL answer in the box, without using any spaces (ex., X=5).

svlad2 [7]

Answer: Y=-1

Explanation: 2y-3=y-4 (add three on both sides)
2y=y-1 (subtract y on both sides)
y=-1

8 0

3 years ago

Is -2-x the same as -2x?

garik1379 [7]

Answer:

yes

Step-by-step explanation:

the variable cannot have a negative symbol, unless it has no number.

4 0

2 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is