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3 years ago

3x/2x simplified into racio

Mathematics

1 answer:

Alik [6]3 years ago

5 0

Answer:

Simplify the expression.

3x2/2

hope this makes sense lol

Step-by-step explanation:

You might be interested in

The ratio of orange juice concentrate to water that Zoe used to make orange

m_a_m_a [10]

Answer:

B.12 ounces

Step-by-step explanation:

step 1

Find out the ounces of orange juice concentrate used yesterday

Let

x ----> ounces of orange juice concentrate used yesterday

y ---> ounces of water used yesterday

we know

that

----> equation A

----> equation B

substitute equation B in equation A

solve for x

step 2

today she wants to make twice as much orange juice.

To find out how much orange juice concentrate she needs, multiply the ounces of orange juice concentrate used yesterday by 2

3 0

3 years ago

Read 2 more answers

An electronics company produces transistors, resistors, and computer chips. Each transistor requires 3 units of copper, 2 unit

Gala2k [10]

Answer:

475 transistors, 25 resistors and 50 computer chips can be produced.

Step-by-step explanation:

Let us consider, p = Number of transistors.

q = Number of resistors.

r = Number of computer chips.

The following three linear equations according to question,

$3\times p + 3\times q + 2\times r = 1600\\2\times p + 1\times q + 1\times r = 1025\\1\times p + 2\times q + 2\times r = 625$

The matrix form of any system, Ax = B

Where, A = Coefficient matrix

B = Constant vector

x = Variable vector

$A = \left[\begin{array}{ccc}3&3&2\\2&1&1\\1&2&2\end{array}\right], x = \left[\begin{array}{ccc}p\\q\\r\end{array}\right], B = \left[\begin{array}{ccc}1600\\1025\\625\end{array}\right]$

The inverse matrix, $A^{-1}$ can be found by using the following formula,

$A^{-1} = \frac{1}{det A}\times (C_{A}) ^{T}$

Where, det A = Determinant of matrix A.

$C_{A}$ = Matrix of cofactors of A

Now, applying this formula to find $A^{-1}$ ;

$det A = \left[\begin{array}{ccc}3&3&2\\2&1&1\\1&2&2\end{array}\right] = 3\times(2-2)-3\times(4-1)+2\times(4-1) = -3$

Here, $det A\neq 0$ , thus the matrix is invertible.

$C_{A} = \left[\begin{array}{ccc}(2-2)&-(4-1)&(4-1)\\-(6-4)&(6-2)&-(6-3)\\(3-2)&-(3-4)&(3-6)\end{array}\right] = \left[\begin{array}{ccc}0&-3&3\\-2&4&-3\\1&1&-3\end{array}\right] \\(C_{A}) ^{T} = \left[\begin{array}{ccc}0&-3&3\\-2&4&-3\\1&1&-3\end{array}\right] ^{T} = \left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right]$

$A^{-1} = \frac{1}{-3}\times\left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right] \\ So, x= \frac{1}{-3} \left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right]\times\left[\begin{array}{ccc}1600\\1025\\625\end{array}\right]= \frac{1}{-3} \left[\begin{array}{ccc}-1425\\-75\\-150\end{array}\right] = \left[\begin{array}{ccc}475\\25\\50\end{array}\right]$

So, p = 475, q = 25, r = 50.

7 0

3 years ago

Write an equation of the line in point-slope form that passes through the given points.

RSB [31]

Answer:

Step-by-step explanation:

point - slope formula: y - y1 = m(x-x1)

we need to find the slope ... or what is m ?

m = rise / run

m = y2 - y1 / x2 - x1

where the given points are (x1,y1) and (x2,y2)

m= 7-5 / 6-5

m = 2 / 1

m =2

now use either point and plug the slope and the point into the point-slope formula

y-5 =2(x-5) :)

5 0

3 years ago

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

kifflom [539]

Looks like we have

$\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k$

which has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2$

By the divergence theorem, the integral of $\vec F$ across $S$ is equal to the integral of $\nabla\cdot\vec F$ over $R$ , where $R$ is the region enclosed by $S$ . Of course, $S$ is not a closed surface, but we can make it so by closing off the hemisphere $S$ by attaching it to the disk $x^2+y^2\le1$ (call it $D$ ) so that $R$ has boundary $S\cup D$ .

Then by the divergence theorem,

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

Compute the integral in spherical coordinates, setting

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

so that the integral is

$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

The integral of $\vec F$ across $S\cup D$ is equal to the integral of $\vec F$ across $S$ plus the integral across $D$ (without outward orientation, so that

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

Then we have

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4$

Finally,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}$

6 0

4 years ago

Which of the sets of ordered pairs represents a function? (1 point) A = {(2, 7), (1, −5), (7, 2), (2, −9)} B = {(5, 3), (−2, −9)

Neporo4naja [7]

Answer:

its B

Step-by-step explanation:

i was taught to do functions by lining up the xs on one side and the ys on the other(if they repeat only put the number once) then draw lines to the pairs. no x value can have two y values.

5 0

3 years ago