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Fynjy0 [20]
3 years ago
14

-48=-5y+8-y Enter only the value of the variable.

Mathematics
2 answers:
Eddi Din [679]3 years ago
4 0
<h3>y = 28 / 3</h3><h3 />

-48=-5y+8-y

First make y positive by * -1.

48 = 5y - 8 + y

5y + y - 8 = 48

6y - 8 = 48

ax = b formula - Take the - 8 to the other side.

6y - 8 + 8 = 48 + 8

6y = 56

x = b formula

Divide by 6:

6y / 6 = 56 / 6

<h3>y = 28 / 3</h3>

Check the answer by substituting it into the original equation:

-48 =-59 (28 / 3) + 8 - (28 / 3)

Lunna [17]3 years ago
3 0
-5( y) +8 -y


y is 14
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Logs are stacked in a pile. The bottom row has 50 logs and next to bottom row has 49 logs. Each row has one less log than the ro
Mars2501 [29]

Answer:

46 logs on the 5th row.

Step-by-step explanation:

Number of logs on the nth row is

n =  50 - (n-1)

 n = 51 - n    (so on the first row we have  51 - 1 = 50 logs).

So on the 5th row we have 51 - 5 = 46 logs.

8 0
3 years ago
Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x
Serggg [28]

I assume C has counterclockwise orientation when viewed from above.

By Stokes' theorem,

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

so we first compute the curl:

\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k

\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k

Then parameterize S by

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

where the z-component is obtained from

1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v

with 0\le u\le\dfrac\pi2 and 0\le v\le\dfrac\pi2.

Take the normal vector to S to be

\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k

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\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}

6 0
3 years ago
Help me with geometry pleasee ( Radom answers will be reported )
IrinaVladis [17]
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8 0
3 years ago
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Masja [62]

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3 0
3 years ago
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hichkok12 [17]

Answer:

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Step-by-step explanation:

To find the angle we need to use the following equation:

d*sin(\theta) = m\lambda

Where:

d: is the separation of the grating

m: is the order of the maximum

λ: is the wavelength

θ: is the angle              

At the first-order maximum (m=1) at 20.0 degrees we have:

\frac{\lambda}{d} = \frac{sin(\theta)}{m} = \frac{sin(20.0)}{1} = 0.342

Now, to produce a second-order maximum (m=2) the angle must be:

sin(\theta) = \frac{\lambda}{d}*m

\theta = arcsin(\frac{\lambda}{d}*m) = arcsin(0.342*2) = 43.2 ^{\circ}

Therefore, the diffraction grating will produce a second-order maximum for the light at 43.2°.    

I hope it helps you!                                                        

6 0
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