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Advocard [28]
3 years ago
13

4x+7+2x+5 solve for X​

Mathematics
2 answers:
Gelneren [198K]3 years ago
8 0

Answer:

x = - 1

Step-by-step explanation:

Let's solve

4x + 7 = 2x + 5

Subtract 2x from both sides.

4x + 7 −2x = 2x + 5 −2x

2x + 7 = 5

Subtract 7 from both sides.

2x + 7 − 7 = 5 −7

2x = −2

Divide both sides by 2.

2x / 2 = −2 / 2

x = −1

dalvyx [7]3 years ago
4 0

Answer:

Simplifying

4x + -7 = -2x + -5

Reorder the terms:

-7 + 4x = -2x + -5

Reorder the terms:

-7 + 4x = -5 + -2x

Solving

-7 + 4x = -5 + -2x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '2x' to each side of the equation.

-7 + 4x + 2x = -5 + -2x + 2x

Combine like terms: 4x + 2x = 6x

-7 + 6x = -5 + -2x + 2x

Combine like terms: -2x + 2x = 0

-7 + 6x = -5 + 0

-7 + 6x = -5

Add '7' to each side of the equation.

-7 + 7 + 6x = -5 + 7

Combine like terms: -7 + 7 = 0

0 + 6x = -5 + 7

6x = -5 + 7

Combine like terms: -5 + 7 = 2

6x = 2

Divide each side by '6'.

x = 0.3333333333

Simplifying

x = 0.3333333333

Step-by-step explanation:

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Find the surface area of the prism.
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Answer:

326m²

Step-by-step explanation:

Find the area of each side and find the total sum from each individual area:

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3 years ago
If f(x)=3x-7, find f(x)=2
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Answer:

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Step-by-step explanation:

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What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po
emmasim [6.3K]
\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA
\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k
\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j

The cross product is

\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j

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5 0
3 years ago
Polygon F has an area of 36 square units. Aimar drew a scaled version of Polygon F and labeled it Polygon G. Polygon G has an ar
Free_Kalibri [48]

Answer:

1/3

Step-by-step explanation:

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

Each side of Polygon FFF was multiplied by a certain value, known as the scale factor , to result in an area that is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

[Show me an example of how scale factor affects area]

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction

 

 

\begin{aligned} A &= \left(l\times\dfrac1{10}\right)\times\left(w\times\dfrac1{10}\right) \\ \\ A&= l\times w\times\dfrac1{10}\times\dfrac1{10} \\ \\ A&= lw \times \left(\dfrac1{10}\right)^2\end{aligned}  

 

 

 

 

 

 

 

 

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction\left(\dfrac1{10}\right)^2  

 

left parenthesis, start fraction, 1, divided by, 10, end fraction, right parenthesis, start superscript, 2, end superscript

Hint #22 / 3

The area of a polygon created with a scale factor of \dfrac1x  

x

1

​  start fraction, 1, divided by, x, end fraction has \left(\dfrac1{x}\right)^2(  

x

1

​  )  

2

left parenthesis, start fraction, 1, divided by, x, end fraction, right parenthesis, start superscript, 2, end superscript the area of the original polygon:

\left(\text{scale factor}\right)^2=\text{fraction of the area the scale copy has}(scale factor)  

2

=fraction of the area the scale copy hasleft parenthesis, s, c, a, l, e, space, f, a, c, t, o, r, right parenthesis, start superscript, 2, end superscript, equals, f, r, a, c, t, i, o, n, space, o, f, space, t, h, e, space, a, r, e, a, space, t, h, e, space, s, c, a, l, e, space, c, o, p, y, space, h, a, s

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF. Let's substitute \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction into the equation to find the scale factor.

\left(\dfrac1{?}\right)^2=\dfrac19(  

?

1

​  )  

2

=  

9

1

​  left parenthesis, start fraction, 1, divided by, question mark, end fraction, right parenthesis, start superscript, 2, end superscript, equals, start fraction, 1, divided by, 9, end fraction

The scale factor is \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction.

Hint #33 / 3

Aimar used a scale factor of \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction to go from Polygon FFF to Polygon GGG.

4 0
3 years ago
Read 2 more answers
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