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andre [41]
3 years ago
14

How do you solve 6g-4=20-2g

Mathematics
2 answers:
suter [353]3 years ago
5 0
6g-4=20-2g
add 2g to both sides
8g-4=20
add 4 to both sides
8g=24
g=3
cluponka [151]3 years ago
5 0
6g-4=20-2g
6g+2g=20+4
8g=24
g=24:8
g=3

<em>Answer:3</em>
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Rationalize <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%20%5Csqrt%7B7%7D%20%2B%203%20%5Csqrt%7B3%7D%20%20%7D%20"
Ivahew [28]
\dfrac{1}{2 \sqrt{7} + 3 \sqrt{3}  }

= \dfrac{1}{2 \sqrt{7} + 3 \sqrt{3} } \times  \dfrac{2 \sqrt{7} - 3 \sqrt{3}}{2 \sqrt{7} - 3 \sqrt{3}}

=\dfrac{2 \sqrt{7} - 3 \sqrt{3}}{(2\sqrt{7})^2 - (3\sqrt{3})^2}

=\dfrac{2 \sqrt{7} - 3 \sqrt{3}}{28 - 27}

=\dfrac{2 \sqrt{7} - 3 \sqrt{3}}{1}

=2 \sqrt{7} - 3 \sqrt{3}

6 0
4 years ago
Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xey + yez + ze
attashe74 [19]

Answer: 6 / √26

Step-by-step explanation:

Given that f(x, y, z) = xe^y + ye^z + ze^x

so first we compute the gradient vector at (0, 0, 0)

Δf ( x, y, z ) = [ e^y + ze^x,  xe^y + e^z,  ye^z + e^x ]

Δf ( 0, 0, 0 ) = [ e⁰ + 0(e)⁰, 0(e)⁰ + e⁰, 0(e)⁰ + e⁰ ] = [ 1+0 , 0+1, 0+1 ] = [ 1, 1, 1 ]

Now we were also given that  V = < 4, 3, -1 >

so ║v║ = √ ( 4² + 3² + (-1)² )

║v║ = √ ( 16 + 9 + 1 )

║v║ = √ 26

It must be noted that "v"  is not a unit vector but since ║v║ = √ 26, the unit vector in the direction of "V" is ⊆ = ( V / ║v║)

so

⊆ =  ( V / ║v║) = [ 4/√26, 3/√26, -1/√26 ]

therefore by equation   D⊆f ( x, y, z ) = Δf ( x, y, z ) × ⊆

D⊆f ( x, y, z ) = Δf ( 0, 0, 0 ) × ⊆ = [ 1, 1, 1 ] × [ 4/√26, 3/√26, -1/√26 ]

= ( 1×4 + 1×3 -1×1 ) / √26

= (4 + 3 - 1) / √26

= 6 / √26

8 0
4 years ago
clara wants to run a total od 5 1/2 miles each lap around the track is 0.25 miles how many laps around the track does clara need
elena-14-01-66 [18.8K]
0.25•4=1
5.5/.25=22

Clara would have to run 22 laps
5 0
3 years ago
What is the linear function equation represented by the graph? Graph of a line on a coordinate plane. The horizontal x-axis rang
salantis [7]
The 2-point form of the equation of a line is
.. y = (y2 -y1)/(x2 -x1)*(x -x1) +y1
For your points (x1, y1) = (0, 3) and (x2, y2) = (3, 5), this is
.. y = (5 -3)/(3 -0)*(x -0) +3
.. y = (2/3)x +3

Written as a function, it is
.. f(x) = (2/3)x +3

8 0
3 years ago
Read 2 more answers
Want to learn symmetry?<br>​
Zanzabum

Answer:

yea

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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