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3 years ago

What is the product property of radicals? Question 4 options:

1 answer:

7 0

If we combine these two things then we get theproduct property of radicals and the quotient property of radicals. These twoproperties tell us that the square root of a product equals the product of the square roots of the factors

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SUB TO YFBG JAY ON YT ASAP RN

Elena L [17]

Answer:

ok on my way to do so!!!!!!

3 0

3 years ago

Given that (- 2, 7) is on the graph of f(x) , find the corresponding point for the function f(x + 4).

AveGali [126]

Answer:

$\boxed{\ the \ corresponding\ point\ is \ (-6,7)\ }$

Step-by-step explanation:

We know that f(-2)=7

x+4 = -2 <=> x = -6

so f(-6+4) = f(-2)=7

then the corresponding point is (-6,7)

3 0

3 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

3 years ago

What is the radius and center of the circle given by the equation (x-5)^{2}+(y+2)^{2} = 81/4

artcher [175]

Answer:

The center is (5, -2) and the radius is 9/2

Step-by-step explanation:

The equation of a circle can be written by

(x-h) ^2 + (y-k)^2 = r^2

where (h,k) is the center and r is the radius

(x-5)^{2}+(y+2)^{2} = 81/4

( (x-5)^{2}+(y- -2)^{2} = (9/2)^2

The center is (5, -2) and the radius is 9/2

6 0

3 years ago

2. Find the product of each of the following:

Kruka [31]

Step-by-step explanation:

$i) \: \frac{3}{5} \times ( \frac{ - 7}{8} ) = - \frac{21}{40} \\ \\ ii) \: \frac{ - 9}{2} \times \frac{5}{4} = - \frac{45}{8} \\ \\ iii) \: \frac{ - 6}{11} \times \frac{5}{3} = - \frac{10}{11} \\ \\ iv) \: \frac{ - 2}{3} \times \frac{6}{7} = - \frac{4}{7} \\$

6 0

3 years ago