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notka56 [123]
3 years ago
13

What should be the first step to solve the equation? 2x + 9(x-1) = 8(2x + 2) -5

Mathematics
1 answer:
tiny-mole [99]3 years ago
6 0

Answer:

It would be to multiply both sides so like you have 9(x-1) = 8(2x + 2)

Step-by-step explanation:

9(x-1) = 8(2x + 2)

9x-9 = 16x+16

and thats the first step

hope this helped :)

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Step-by-step explanation:

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Step-by-step explanation:

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2.7·6.2–9.3·1.2+6.2·9.3–1.2·2.7 not pemdas
antiseptic1488 [7]

<em></em>

<em>60</em>

<em>See steps</em>

<em>Step by Step Solution:</em>

<em>More Icon</em>

<em>Reformatting the input :</em>

<em>Changes made to your input should not affect the solution:</em>

<em />

<em>(1): "2.7" was replaced by "(27/10)". 8 more similar replacement(s)</em>

<em />

<em>STEP</em>

<em>1</em>

<em>:</em>

<em>            27</em>

<em> Simplify   ——</em>

<em>            10</em>

<em>Equation at the end of step</em>

<em>1</em>

<em>:</em>

<em>     27 62   93 12    62 93    12 27</em>

<em>  (((——•——)-(——•——))+(——•——))-(——•——)</em>

<em>     10 10   10 10    10 10    10 10</em>

<em>STEP</em>

<em>2</em>

<em>:</em>

<em>            6</em>

<em> Simplify   —</em>

<em>            5</em>

<em>Equation at the end of step</em>

<em>2</em>

<em>:</em>

<em>     27 62   93 12    62 93    6 27</em>

<em>  (((——•——)-(——•——))+(——•——))-(—•——)</em>

<em>     10 10   10 10    10 10    5 10</em>

<em>STEP</em>

<em>3</em>

<em>:</em>

<em>            93</em>

<em> Simplify   ——</em>

<em>            10</em>

<em>Equation at the end of step</em>

<em>3</em>

<em>:</em>

<em>     27 62   93 12    62 93   81</em>

<em>  (((——•——)-(——•——))+(——•——))-——</em>

<em>     10 10   10 10    10 10   25</em>

<em>STEP</em>

<em>4</em>

<em>:</em>

<em>            31</em>

<em> Simplify   ——</em>

<em>            5 </em>

<em>Equation at the end of step</em>

<em>4</em>

<em>:</em>

<em>     27 62   93 12    31 93   81</em>

<em>  (((——•——)-(——•——))+(——•——))-——</em>

<em>     10 10   10 10    5  10   25</em>

<em>STEP</em>

<em>5</em>

<em>:</em>

<em>            6</em>

<em> Simplify   —</em>

<em>            5</em>

<em>Equation at the end of step</em>

<em>5</em>

<em>:</em>

<em>     27 62   93 6   2883  81</em>

<em>  (((——•——)-(——•—))+————)-——</em>

<em>     10 10   10 5    50   25</em>

<em>STEP</em>

<em>6</em>

<em>:</em>

<em>            93</em>

<em> Simplify   ——</em>

<em>            10</em>

<em>Equation at the end of step</em>

<em>6</em>

<em>:</em>

<em>     27 62   93 6   2883  81</em>

<em>  (((——•——)-(——•—))+————)-——</em>

<em>     10 10   10 5    50   25</em>

<em>STEP</em>

<em>7</em>

<em>:</em>

<em>            31</em>

<em> Simplify   ——</em>

<em>            5 </em>

<em>Equation at the end of step</em>

<em>7</em>

<em>:</em>

<em>     27   31     279     2883     81</em>

<em>  (((—— • ——) -  ———) +  ————) -  ——</em>

<em>     10   5      25       50      25</em>

<em>STEP</em>

<em>8</em>

<em>:</em>

<em>            27</em>

<em> Simplify   ——</em>

<em>            10</em>

<em>Equation at the end of step</em>

<em>8</em>

<em>:</em>

<em>     27   31     279     2883     81</em>

<em>  (((—— • ——) -  ———) +  ————) -  ——</em>

<em>     10   5      25       50      25</em>

<em>STEP</em>

<em>9</em>

<em>:</em>

<em>Calculating the Least Common Multiple</em>

<em> 9.1    Find the Least Common Multiple</em>

<em />

<em>      The left denominator is :       50 </em>

<em />

<em>      The right denominator is :       25 </em>

<em />

<em>        Number of times each prime factor</em>

<em>        appears in the factorization of:</em>

<em> Prime </em>

<em> Factor   Left </em>

<em> Denominator   Right </em>

<em> Denominator   L.C.M = Max </em>

<em> {Left,Right} </em>

<em>2 1 0 1</em>

<em>5 2 2 2</em>

<em> Product of all </em>

<em> Prime Factors  50 25 50</em>

<em />

<em>      Least Common Multiple:</em>

<em>      50 </em>

<em />

<em>Calculating Multipliers :</em>

<em> 9.2    Calculate multipliers for the two fractions</em>

<em />

<em />

<em>    Denote the Least Common Multiple by  L.C.M </em>

<em>    Denote the Left Multiplier by  Left_M </em>

<em>    Denote the Right Multiplier by  Right_M </em>

<em>    Denote the Left Deniminator by  L_Deno </em>

<em>    Denote the Right Multiplier by  R_Deno </em>

<em />

<em>   Left_M = L.C.M / L_Deno = 1</em>

<em />

<em>   Right_M = L.C.M / R_Deno = 2</em>

<em />

<em />

<em>Making Equivalent Fractions :</em>

<em> 9.3      Rewrite the two fractions into equivalent fractions</em>

<em />

<em>Two fractions are called equivalent if they have the same numeric value.</em>

<em />

<em>For example :  1/2   and  2/4  are equivalent,  y/(y+1)2   and  (y2+y)/(y+1)3  are equivalent as well.</em>

<em />

<em>To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.</em>

<em />

<em>   L. Mult. • L. Num.      837</em>

<em>   ——————————————————  =   ———</em>

<em>         L.C.M             50 </em>

<em />

<em>   R. Mult. • R. Num.      279 • 2</em>

<em>   ——————————————————  =   ———————</em>

<em>         L.C.M               50   </em>

<em>Adding fractions that have a common denominator :</em>

<em> 9.4       Adding up the two equivalent fractions</em>

<em>Add the two equivalent fractions which now have a common denominator</em>

<em />

<em>Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:</em>

<em />

<em> 837 - (279 • 2)     279</em>

<em> ———————————————  =  ———</em>

<em>       50            50 </em>

<em>Equation at the end of step</em>

<em>9</em>

<em>:</em>

<em>   279    2883     81</em>

<em>  (——— +  ————) -  ——</em>

<em>   50      50      25</em>

<em>STEP</em>

<em>10</em>

<em>:</em>

<em>Adding fractions which have a common denominator</em>

<em> 10.1       Adding fractions which have a common denominator</em>

<em>Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:</em>

<em />

<em> 279 + 2883     1581</em>

<em> ——————————  =  ————</em>

<em>     50          25 </em>

<em>Equation at the end of step</em>

<em>10</em>

<em>:</em>

<em>  1581    81</em>

<em>  ———— -  ——</em>

<em> </em>  25     25

STEP

11

:

Adding fractions which have a common denominator

11.1       Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

1581 - (81)     60

———————————  =  ——

    25          1

Final result :

 60

7 0
3 years ago
An alligator is 2.3 yards long what is the length of the alligator in inches
grin007 [14]
The alligator is 82.2 inches long
3 0
3 years ago
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Use a surface integral to find the surface area of the portion of the sphere xUse a surface integral to find the surface area of t
brilliants [131]

Parameterize this surface (call it S) by

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

with 0\le u\le2\pi and 0\le v\le\dfrac\pi4. The limits on u should be obvious. We find the upper limit for v by solving for v where the sphere and cone intersect:

\begin{cases}x^2+y^2+z^2=1\\z=\sqrt{x^2+y^2}\end{cases}\implies x^2+y^2=\dfrac12

\implies(\cos u\sin v)^2+(\sin u\sin v)^2=\dfrac12

\implies\sin^2v=\dfrac12

\implies\sin v=\dfrac1{\sqrt2}\implies v=\dfrac\pi4

Take the normal vector to S to be

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

(orientation does not matter here)

Then the area of S is

\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{\pi/4}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv

=\displaystyle2\pi\int_0^{\pi/4}\sin v\,\mathrm dv=\boxed{(2-\sqrt2)\pi}

4 0
3 years ago
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