Match the numbers to their equivalent alternate form.

Mathematics

1 answer:

ValentinkaMS [17]3 years ago

5 0

Answer:

6558879790645768543

Step-by-step explanation:

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Rude took a test and earned a 92%. Camilo earned an 88% on the same test. If Rudy answered 46 questions correctly, how many more

g100num [7]

Rudy answered 2 more questions correctly
...........
46x2=92
44x2=88
...........
92-88=4

7 0

4 years ago

Anyone good with algebra 2??

sweet [91]

A discontinuity is a point that cannot exist because the x-coordinate would cause a problem in the equation. If you have a polynomial in the denominator, you must find which values of x would cause the polynomial in the denominator to evaluate to zero. Since division by zero is undefined, that would cause a discontinuity.

Let's look at your function.

$f(x) = \dfrac{x^2 - 16}{6x - 24}$

$x^2 - 16$ is in the numerator. It is defined for every value of x. There is no problem there.

$6x - 24$ is in the denominator. This is a function defined for every value of x, but since it is in the denominator, we must exclude the x-value that would cause this polynomial to evaluate to zero.

We set it equal to zero and solve the equation for x.

$6x - 24 = 0$

$6x = 24$

$x = 4$

For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).

The answer is:

$x \ne 4$

6 0

3 years ago

Evaluate the spherical coordinate integral

expeople1 [14]

Rewrite the equations of the given boundary lines:

y = -x + 1 ==> x + y = 1

y = -x + 4 ==> x + y = 4

y = 2x + 2 ==> -2x + y = 2

y = 2x + 5 ==> -2x + y = 5

This tells us the parallelogram in the x-y plane corresponds to the rectangle in the u-v plane with 1 ≤ u ≤ 4 and 2 ≤ v ≤ 5.

Compute the Jacobian determinant for this change of coordinates:

$J=\begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}1&1\\-2&1\end{bmatrix}\implies|\det J|=3$

Rewrite the integrand:

$-3x+4y=-3\cdot\dfrac{u-v}3+4\cdot\dfrac{2u+v}3=\dfrac{5u+7v}3$

The integral is then

$\displaystyle\iint_R(-3x+4y)\,\mathrm dx\,\mathrm dy=3\iint_{R'}\frac{5u+7v}3\,\mathrm du\,\mathrm dv=\int_2^5\int_1^45u+7v\,\mathrm du\,\mathrm dv=\boxed{333}$