All categories

3 years ago

What is 5.74 in radical form

Mathematics

1 answer:

svetlana [45]3 years ago

3 0

Answer:

x^4/7

Step-by-step explanation:

You might be interested in

The fuse are building a new home one of the bedroom closet has an area of 48 square feet and a perimeter of 32 ft what are the l

olya-2409 [2.1K]

Answer: The length is 12 feet and the width is 4 feet.

Step-by-step explanation:

Let L represent the length of the bedroom closet.

Let W represent the width of the bedroom closet..

The formula for determining the perimeter of a rectangle is expressed as

Perimeter = 2(L + W)

The perimeter of the bedroom closet is 32 ft. This means that

2(L + W) = 32

Dividing through by 2, it becomes

L + W = 32/2

L + W = 16

The area of the bedroom closet is 48 ft². This means that

LW = 48 - - - - - - - -- - - - - - 1

Substituting L = 16 - W into equation 1, it becomes

W(16 - W) = 48

16W - W² = 48

W² - 16W + 48 = 0

W² - 12W - 4W + 48 = 0

W(W - 12) - 4(W - 12) = 0

W - 12 = 0 or W - 4 = 0

W = 12 or W = 4

L = 16 - 4 = 12

3 0

4 years ago

Suppose a > 0 is constant and consider the parameteric surface sigma given by r(phi, theta) = a sin(phi) cos(theta)i + a sin(

Gnom [1K]

$\Sigma$ should have parameterization

$\vec r(\varphi,\theta)=a\sin\varphi\cos\theta\,\vec\imath+a\sin\varphi\sin\theta\,\vec\jmath+a\cos\varphi\,\vec k$

if it's supposed to capture the sphere of radius $a$ centered at the origin. ( $\sin\theta$ is missing from the second component)

a. You should substitute $x=a\sin\varphi\cos\theta$ (missing $\cos\theta$ this time...). Then

$x^2+y^2+z^2=(a\sin\varphi\cos\theta)^2+(a\sin\varphi\sin\theta)^2+(a\cos\varphi)^2$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\cos^2\theta+\sin^2\varphi\sin^2\theta+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\left(\cos^2\theta+\sin^2\theta\right)+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2$

as required.

b. We have

$\vec r_\varphi=a\cos\varphi\cos\theta\,\vec\imath+a\cos\varphi\sin\theta\,\vec\jmath-a\sin\varphi\,\vec k$

$\vec r_\theta=-a\sin\varphi\sin\theta\,\vec\imath+a\sin\varphi\cos\theta\,\vec\jmath$

$\vec r_\varphi\times\vec r_\theta=a^2\sin^2\varphi\cos\theta\,\vec\imath+a^2\sin^2\varphi\sin\theta\,\vec\jmath+a^2\cos\varphi\sin\varphi\,\vec k$

$\|\vec r_\varphi\times\vec r_\theta\|=a^2\sin\varphi$

c. The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=a^2\int_0^\pi\int_0^{2\pi}\sin\varphi\,\mathrm d\theta\,\mathrm d\varphi$

You don't need a substitution to compute this. The integration limits are constant, so you can separate the variables to get two integrals. You'd end up with

$\displaystyle\iint_\Sigma\mathrm dS=4\pi a^2$

# # #

Looks like there's an altogether different question being asked now. Parameterize $\Sigma$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $\sqrt2\le u\le\sqrt6$ and $0\le v\le2\pi$ . Then

$\|\vec s_u\times\vec s_v\|=u\sqrt{1+4u^2}$

The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=\int_0^{2\pi}\int_{\sqrt2}^{\sqrt6}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

The integrand doesn't depend on $v$ , so integration with respect to $v$ contributes a factor of $2\pi$ . Substitute $w=1+4u^2$ to get $\mathrm dw=8u\,\mathrm du$ . Then

$\displaystyle\iint_\Sigma\mathrm dS=\frac\pi4\int_9^{25}\sqrt w\,\mathrm dw=\frac{49\pi}3$

# # #

Looks like yet another different question. No figure was included in your post, so I'll assume the normal vector points outward from the surface, away from the origin.

Parameterize $\Sigma$ by

$\vec t(u,v)=u\,\vec\imath+u^2\,\vec\jmath+v\,\vec k$

with $-1\le u\le1$ and $0\le v\le 2$ . Take the normal vector to $\Sigma$ to be

$\vec t_u\times\vec t_v=2u\,\vec\imath-\vec\jmath$

Then the flux of $\vec F$ across $\Sigma$ is

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^2\int_{-1}^1(u^2\,\vec\jmath-uv\,\vec k)\cdot(2u\,\vec\imath-\vec\jmath)\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-\int_0^2\int_{-1}^1u^2\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-2\int_{-1}^1u^2\,\mathrm du=-\frac43$

If instead the direction is toward the origin, the flux would be positive.

8 0

4 years ago

last summer at Camp okey dokey fun okey the ratio of the number of boy campers to the number of girl campers was 8 to 7 if there

Lena [83]

8x = boys
7x = girls

8x + 7x = 195
15x = 195
/15
x = 13

8x = 8 • 13 = 104
7x = 7 • 13 = 91

104 boys and 91 girls

4 0

4 years ago

lathasha made 128 ounces of punch how many cups of punch did latasha make? A 4 cups B 8 cups C 16 cups D 32 please help

Vera_Pavlovna [14]

128 ounces is equal to 16 cups

4 0

3 years ago

Read 2 more answers

How do i find x (time) when it's an exponential function? please help

allochka39001 [22]

Using exponentials inverse function (logarithms )
Or common base method which is not always easily applicable

ln600 = ln2100 + xln0.5

x = ln(600/2100) / ln0.5 = 1.81

6 0

3 years ago