All categories

3 years ago

18 ➗ (-4 1/2) with explanation

Mathematics

1 answer:

slamgirl [31]3 years ago

8 0

Answer:

-4

Step-by-step explanation:

If anything the first thing you would want to do is covert your fraction so a decimal. (-4 1/2) as a decimal is -4.50. From there it would be easier to divide. to get an answer of -4

You might be interested in

Please help with last 2 domain and range

Strike441 [17]

Answer:

See below for answers and explanations

Step-by-step explanation:

Top left: Since y can't be greater than 0 but is equal to 0, then the range is (-∞,0] and the domain is (-∞,∞) since there are no domain restrictions

Top right: Since both x and y have no restrictions, then the domain is (-∞,∞) and the range is (-∞,∞)

Bottom left: Since y cannot be less than 2 but equal to it, and x holds no domain restrictions, then the domain is (-∞,∞) and the range is [2,∞)

Bottom right: Since both x and y have no restrictions, then the domain is (-∞,∞) and the range is (-∞,∞)

7 0

3 years ago

Help would be greatly appreciated :)

Anton [14]

The bottle of 100 would be lowest. $8.13 ÷ 100 lozenges = $0.08 per lozenge

7 0

3 years ago

Ruby has a loss of $45 in her business. How will she enter the loss in her account book?

user100 [1]

-45 is the correct answer

9 0

3 years ago

512, 256, 128, 64, __, __, __, ...

Cerrena [4.2K]

The complete pattern is: 512, 256, 128, 64, 32, 16, 8, ( and extended ) 4, 2, 1!

I do this pattern in my head literally all the time!

4 0

3 years ago

Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

LUCKY_DIMON [66]

Make a change of coordinates:

$u(x,y)=xy$
$v(x,y)=\dfrac xy$

The Jacobian for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

and has a determinant of

$\det\mathbf J=-\dfrac{2x}y$

Note that we need to use the Jacobian in the other direction; that is, we've computed

$\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}$

but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

$\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}$

we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
$=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2$

8 0

3 years ago

18 ➗ (-4 1/2) with explanation​

18 ➗ (-4 1/2) with explanation