Solve using elimination. 9x − 9y = –18 10x − 9y =

How many one-thirds are in one-sixth?

andrew11 [14]

there are (1/6)/(1/3) one-third in one sixth

Step-by-step explanation:

that means

no. of one-third is 1/2

4 0

3 years ago

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Find the reference angle of <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20-%20%5Cpi%7D%7B7%7D%20" id="TexFormula1" title

Semenov [28]

Answer:

π7

Step-by-step explanation:

since is in the first quadrant, the reference angle is π7

3 0

3 years ago

A bicycle computer or cyclometer uses a magnet counter that records each wheel rotation to calculate the bike’s total distance t

CaHeK987 [17]

The simplest and probably the best way to understand this problem is to make up a problem that obeys what you have been given. It doesn't have to be realistic. It just has to obey the conditions. Let us suppose that you thought the diameter of the tire is 1 yard. That would mean the circumfrence is pi * d

C = 3.14 * 1

That would mean that the circumference is 3.14 yards. It would also mean that you would have to have the wheel turn 1760 yards / /3.14 yards / revolution which is about 561 revolutions / mile. So the way I have set up the problem, my equation is d = 561 * R where R is the number of revolutions.

Now let's see what happens when you say "O my Goodness, the wheel diameter is really 32 inches" which 0.8888888 yards what happens now?

Now you still have to go 1760 yards How many revolutions is that?

C = pi * d

C = 3.14 * 0.88888888

C = 2.79111 yards

How many revolutions does it take to 1760 yards.

R = 1760 // 2.78111 yards / revolution

R = 631 revolutions / mile. What happened?

Your constant goes up if the wheel diameter goes down. Think about this. Do you ride a bicycle? I do. It makes perfect sense to me that if the wheel is small, it will have to turn more often to go a mile. No matter where that 0.00125 comes from or how it was derived, the constant will have to go up if the wheel gets smaller.

6 0

3 years ago

PLEASE HELP TIMER!!!!!!!!!!!!!

xz_007 [3.2K]

Answer:

Answer: -3/2

Step-by-step explanation:

#KEEPSAFE

#GOODLUCK

#STUDYWELL

5 0

1 year 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

3 years ago

Solve using elimination. 9x − 9y = –18 10x − 9y = –13