Hi there
The formula you meant
V (t)=27500 (0.84)^t
V (10)=27,500×(0.84)^(10)
V (10)=4,809.8 Round your answer to get 4810 ...this isthe value after
10 years

the initial value of the car is
27500

Hope it helps

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

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

2 years ago

Does anyone know 1/4 x ? = -2?

SSSSS [86.1K]

Answer:

-8

Step-by-step explanation:

1/4 x = -2

You want the value of x. x is being multiplied by 1/4. To get x alone on the left side, you need to divide both sides by 1/4. Dividing by 1/4 is the same as multiplying by 4.

Multiply both sides of the equation by 4.

4 * 1/4 x = -2 * 4

4 * 1/4 = 1, so on the left side you get 1x which is just x.

x = 8

Answer: -8

8 0

2 years ago

Read 2 more answers

A 1 ​ =7 and a_n=a_{n-1}+4a n ​ =a n−1 ​ +4

A 1 =7 and a_n=a_{n-1}+4a n =a n−1 +4