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what is 2(x+4)-5(x-3)=32

Need help in solving x in this equation

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Three rules of exponents - A complete course in algebra

... Section 2 Three Rules of Exponents. Back to Section 1. ... Multiply 3x 2 · 4x 5 ... 5 = 32a 10 b 15 : d) (xy 3 z 5) 2 = x 2 y 6 z 10 : e) (5x 2 y 4) ...

What is (X^(2)) * (X^(1/2))? - Quora

$x^{5/2} = (x^{1/2})^5 = \sqrt{x}^5 ... [math] \sqrt{4}^5 = 2^5 = 32$. You can verify that that's the same as plugging 4 into your original expression:

5 Ways to Solve for X - wikiHow

How to Solve for X. There are a number of ways to solve for x, ... 2 2 (4+3)+ 9 - 5 = 32; 2 2 (7) + 9 - 5 = 32; 4(7) + 9 - 5 = 32; 28 + 9 - 5 = 32; 37 - 5 = 32; 32 ...

Did you solve it? Complete the equation 10 9 8 7 6 5 4 3 2 1 ...

Did you solve it? Complete the equation 10 9 8 7 6 5 4 3 2 1 = 2016 ... (32 and 9) out of (10 9 8) and (6 5 4 3 2 1). If this worked I’d have (10 9 8) x 7 x ...

(4x-5)^2 = -32 - Math Central

(4x-5)^2 = -32 . Hi Tracey, You could ... Whenever I do this kind of algebra, I like to check my answer by putting it in for x in the original equation: (4x-5) 2 =(4 ...

How to solve for $x$ in $x^2=16^x$ - Quora

$$$\begin{split}x^2&=16^x\\ \implies |x|&=4^x\end{split}$$\tag*{}$ [math]\text{Case I:}\\ \begin{split}x&=4^x ...

Untitled Document [www.math.brown.edu]

(x 2 + 4x - 1)(x 3 - 2x 2 + 4) = x 5 - 2x 4 + 4x 2 + 4x 4 - 8x 3 + 16x - x 3 + 2x 2 - 4 (x 2 + 4x - 1)(x 3 - 2x 2 + 4) = x 5 + 2x 4 - 9x 3 + 6x 2 + 16x - 4. Factoring ...

15/32 in. x 4 ft. x 8 ft. 3-Ply RTD Sheathing - The Home Depot

15/32 in. x 4 ft. x 8 ft. 3-Ply RTD Sheathing Questions - page 2; y_2017, m_6, d_25, h_6CST; ... The Home Depot Canada; The Home Depot México; Blinds.com; Pro Referral

Solving an Equation - Avvanta

Solving an Equation Here is a typical equation: 3X - 5 = 27 ... X = 6: Because 3 times 6 = 18: X/4 = 5: If you divide my number by 4, you get 5. What is my number? X ...

1/32 = .03125 1/16 = .0625 3/32 = .09375 1/8 = .125 5/32 ...

5/32 = .15625 3/16 = .1875 7/32 = .21875 1/4 = .25 9/32 = .28125 5/16 = .3125 11/32 = .34375 3/8 = .375 13/32 = .40625 7/16 = .4375 15/32 = .46875 1/2 = .5 17/32 = .53125

f(x)=2x4-x3

f(x) = 2x^4 - x^3 f'(x) = 8x^3 - 3x^2 Setting the derivative to zero, we have: f'(x) = 0 8x^3 - 3x^2 = 0 x^2(8x - 3) = 0 x^2 = 0 or 8x - 3 = 0 x = 0 or x = 3 / 8 Thus, the critical values we have are x = 0 and x = 3 / 8. Now, we will draw a number line with 0 and 3/8 on it. By random substitution, we will realise fo any number less than 0, f'(x) will be more than zero. Similarly, for any number more than 3/8, f'(x) will also be more than zero. This implies that the function is increasing on (-infinity, 0) and (3/8, infinity). However, by using the similar random substitution, we will realise that for any value between 0 and 3/8, f'(x) will be negative. Thus, the function is decreasing on the interval (0, 3/8).

find all rational and non-rational zeros P(x)=2x^4 + 5x^3 − 5x^2 − 5x + 3

P(x) = 2x4 + 5x3 - 5x2 - 5x +3 ⇒P(1) = 2 + 5 - 5 - 5 + 3 = 0 ⇒ x - 1 is a factor ⇒P(-1) = 2 - 5 - 5 + 5 + 3 = 0 ⇒ x + 1 is a factor ⇒(2x4 + 5x3 - 5x2 - 5x +3)/(x2 - 1) = 2x2 + 5 x - 3 ⇒ (x2 - 1)(2x2 + 5 x - 3) = (x + 1)(x - 1)(2x - 1)(x + 3) = 0 ⇒ x = -3, -1, ½, 1

how do I fiqure this out 2x4^3x=8x/16

The right-hand side can be simplified to x/2. By doubling, we can get rid of the fraction: 4x4^3x=x. Assuming this is meant to be: 4x(4^3x)=x then we have x(4*4^3x-1)=0 and x=0 is a solution. The other solution is 4^3x=1/4. Take logs to the base 4: 3x=-1 so x=-1/3. Alternatively, we can write 4*4^3x-1 as 4^(3x+1)-1, so 4^(3x+1)=1 and 3x+1=0, giving x=-1/3.

BET you cant answer this! ~~~HELP~~~

None of the graphs match the pictures. The first equation contains the point (0,42), which does not appear on any of the graphs. The second equation contains the point (1,9), which does not appear on any of the graphs. The third equation contains the point (0,-8), which does not appear on any of the graphs. The fourth equation contains the point (1,-10), which does not appear on any of the graphs. The fifth equation contains the point (1,1), which does not appear on any of the graphs. The sixth equation contains the point (0,1), which does not appear on any of the graphs.

Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0.

Question: Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0. The tangent plane to the surface z=f(x,y)=xy^3+8/y is given by 2x+7y+2z=D Where D is some constant. The equation of the tangent plane to the surface f(x,y), at the point (x_0 y_0 ) can be written as z=f(x_0,y_0 ) + f_x (x_0 y_0 )∙(x-x_0 ) + f_(y ) (x_0,y_0 )∙(y-y_0 ) Where (x_0,y_0) is some point on that plane andf_x (x_0 y_0 ) is the slope of the tangent line to the surface at that point in the x-direction and similarly for f_(y ) (x_0,y_0 ). We have f(x,y)=xy^3+8/y. f_x=y^3,  f_y=3xy^2-8/y^2 . So, f_x (x_0 y_0 )=y_0^3, f_(y ) (x_0,y_0 )=3x_0 y_0^2-8/(y_0^2 ), f(x_0,y_0 )=x_0 y_0^3+8/y_0 . Substituting for the above into the equation of the tangent plane, z = x_0 y_0^3+8/y_0 + y_0^3∙(x-x_0 ) + (3x_0 y_0^2 - 8/(y_0^2 ))∙(y-y_0 ) z = x_0 y_0^3+8/y_0 + y_0^3 x - x_0 y_0^3 + (3x_0 y_0^2 - 8/(y_0^2 ))y - 3x_0 y_0^3 + 8/y_0 z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {x_0 y_0^3 + 8/y_0 - x_0 y_0^3 - 3x_0 y_0^3 + 8/y_0 } z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {16/y_0 - 3x_0 y_0^3 } -2y_0^3 x - 2(3x_0 y_0^2 - 8/(y_0^2 ))y + 2z = 32/y_0 - 6x_0 y_0^3 We have just written down the equation of the tangent plane, which has also been written down as 2x+7y+2z=D Comparing the two forms of these equations, -2y_0^3 = 2     ----------------------------------- (1) 2(3x_0 y_0^2 - 8/(y_0^2 )) = 7     ---------- (2) 32/y_0 - 6x_0 y_0^3 = D   ------------------- (3) From (1), y_0 = -1 Substituting for y_0=-1 into (2), we get x_0 = 3/2. Substituting for x_0 and y_0 into (3), we get D = -23, and z = f(x_0,y_0 ) = (3/2) (-1)^3 + 8/(-1) = (-9.5) The equation of the tangent plane is thus 2x+7y+2z=-23 And this plane is tangential to the surface f(x,y) at one point only, which is: (1.5,-1,-9.5)

what are the possible combination of making 30 with 1, 3, 5, 7, 9

I started off by figuring how many ways to use nines: 30   3 nines, 27 total   2 nines, 18 total   1 nine, 9 total   0 nines, 0 total I then figured how many ways to use sevens: 30   3 nines, 27 total     0 sevens, 27 total   2 nines, 18 total     1 seven, 25 total     0 sevens, 18 total   1 nine, 9 total     3 sevens, 30 total     2 sevens, 23 total     1 seven, 16 total     0 sevens, 9 total   0 nines, 0 total     4 sevens, 28 total     3 sevens, 21 total     2 sevens, 14 total     1 seven, 7 total     0 sevens, 0 total I then figured how many ways to use fives, then threes, ending up with this: 30   3 nines, 27 total     0 sevens, 27 total       0 fives, 27 total         1 three, 30 total         0 threes, 27 total   2 nines, 18 total     1 seven, 25 total       1 five, 30 total         0 threes, 30 total       0 fives, 25 total         1 three, 28 total         0 threes, 25 total     0 sevens, 18 total       2 fives, 28 total         0 threes, 28 total       1 five, 23 total         2 threes, 29 total         1 three, 26 total         0 threes, 23 total       0 fives, 18 total         4 threes, 30 total         3 threes, 27 total         2 threes, 24 total         1 three, 21 total         0 three, 18 total   1 nine, 9 total ​    3 sevens, 30 total       0 fives, 30 total         0 threes, 30 total     2 sevens, 23 total       1 five, 28 total         0 threes, 28 total       0 fives, 23 total         2 threes, 29 total         1 three, 26 total         0 threes, 23 total     1 seven, 16 total       2 fives, 26 total         1 three, 29 total         0 threes, 26 total       1 five, 21 total         3 threes, 30 total         2 threes, 27 total         1 three, 24 total         0 threes, 21 total       0 fives, 16 total         4 threes, 28 total         3 threes, 25 total         2 threes, 22 total         1 three, 19 total     0 threes, 16 total     0 sevens, 9 total       4 fives, 29 total     0 threes, 29 total       3 fives, 24 total     2 threes, 30 total     1 three, 27 total     0 threes, 24 total       2 fives, 19 total     3 threes, 28 total     2 threes, 25 total     1 three, 22 total     0 threes, 19 total       1 five, 14 total     5 threes, 29 total     4 threes, 26 total         3 threes, 23 total     2 threes, 20 total     1 three, 17 total     0 threes, 14 total       0 fives, 9 total     7 threes, 30 total     6 threes, 27 total     5 threes, 24 total     4 threes, 21 total     3 threes, 18 total     2 threes, 15 total     1 three, 12 total     0 threes, 9 total   0 nines, 0 total     4 sevens, 28 total       0 fives, 28 total     0 threes, 28 total     3 sevens, 21 total       1 five, 26 total     1 three, 29 total     0 threes, 26 total       0 fives, 21 total     3 threes, 30 total     2 threes, 27 total     1 three, 24 total     0 threes, 21 total     2 sevens, 14 total       3 fives, 29 total     0 threes, 29 total       2 fives, 24 total     2 threes, 30 total     1 three, 27 total     0 threes, 24 total       1 five, 19 total     3 threes, 28 total     2 threes, 25 total     1 three, 22 total     0 threes, 19 total       0 fives, 14 total     5 threes, 29 total     4 threes, 26 total     3 threes, 23 total     2 threes, 20 total     1 three, 17 total     0 threes, 14 total     1 seven, 7 total       4 fives, 27 total     1 three, 30 total     0 threes, 27 total       3 fives, 22 total     2 threes, 28 total     1 three, 25 total     0 threes, 22 total       2 fives, 17 total     4 threes, 29 total     3 threes, 26 total     2 threes, 23 total     1 three, 20 total     0 threes, 17 total       1 five, 12 total     6 threes, 30 total     5 threes, 27 total     4 threes, 24 total     3 threes, 21 total     2 threes, 18 total     1 three, 15 total     0 threes, 12 total       0 fives, 7 total     7 threes, 28 total     6 threes, 25 total     5 threes, 22 total     4 threes, 19 total     3 threes, 16 total     2 threes, 13 total     1 three, 10 total     0 threes, 7 total     0 sevens, 0 total       6 fives, 30 total     0 threes, 30 total       5 fives, 25 total     1 three, 28 total     0 threes, 25 total       4 fives, 20 total     3 threes, 29 total     2 threes, 26 total     1 three, 23 total     0 threes, 20 total       3 fives, 15 total     5 threes, 30 total     4 threes, 27 total     3 threes, 24 total     2 threes, 21 total     1 three, 18 total     0 threes, 15 total       2 fives, 10 total     6 threes, 28 total     5 threes, 25 total     4 threes, 22 total     3 threes, 19 total     2 threes, 16 total     1 three, 13 total     0 threes, 10 total       1 five, 5 total     8 threes, 29 total     7 threes, 26 total     6 threes, 23 total     5 threes, 20 total     4 threes, 17 total     3 threes, 14 total     2 threes, 11 total     1 three, 8 total     0 threes, 5 total       0 fives, 0 total     10 threes, 30 total     9 threes, 27 total     8 threes, 24 total     7 threes, 21 total     6 threes, 18 total     5 threes, 15 total     4 threes, 12 total     3 threes, 9 total     2 threes, 6 total     1 three, 3 total     0 threes, 0 total (more to follow)

What is inverse

If B= ( b 0 0 ) ( 0 b 0 ) = bI, where I is the identity matrix, ( 0 0 b ) then B^7= ( b^7 0 0 ) ( 0 b^7 0 ) = ( 0 0 b^7 ) So  ( b^7 0 0 )   ( 3b 0 0 )    ( 1 0 0 )    ( 0 0 0 ) ( 0 b^7 0 ) - ( 0 3b 0 ) + ( 0 1 0 ) = ( 0 0 0 ). ( 0 0 b^7 )   ( 0 0 3b )     ( 0 0 1 )    ( 0 0 0 ) b^7-3b+1=0 applies right across the matrices and this equation has 3 real roots which we can call b1, b2, b3, where b1, b2, b3 =-1.249, 0.333, 1.133 approx. B=bI so the inverse of B=B'=I÷b, where b can be any one of b1, b2, b3. B'= ( 1/b 0 0 )    ( -0.801 0 0 )   ( 3.003 0 0 )  ( 0.8826 0 0 ) ( 0 1/b 0 ) = ( 0 -0.801 0 ),  ( 0 3.003 0 ), ( 0 0.8826 0 ) ( 0 0 1/b)     ( 0 0 -0.801)    ( 0 0 3.003 )  ( 0 0 0.8826 ).

(1) I am assuming this is not a calculus question where dx, dy and dz have a different meaning. Variable d can be removed as a common factor: ax/((b-c)yz)=by/((c-a)zx)=cz/((a-b)xy) Take the first pair and remove common factor z: ax/((b-c)y)=by/((c-a)x); ax^2(c-a)=by^2(b-c); ax^2=by^2(b-c)/(c-a); by^2=ax^2(c-a)/(b-c) The second pair have common factor x: by^2(a-b)=cz^2(c-a); by^2=cz^2(c-a)/(a-b); cz^2=by^2(a-b)/(c-a) First and last have common factor y: ax^2(a-b)=cz^2(b-c); ax^2=cz^2(b-c)/(a-b); cz^2=ax^2(a-b)/(b-c) We have a pair of equations for each of the three quantities ax^2, by^2, cz^2. So, ax^2=by^2(b-c)/(c-a)=cz^2(b-c)/(a-b), by^2/(c-a)=cz^2/(a-b) or (A) by^2/(cz^2)=(c-a)/(a-b) by^2=ax^2(c-a)/(b-c)=cz^2(c-a)/(a-b), ax^2/(b-c)=cz^2/(a-b) or (B) ax^2/(cz^2)=(b-c)/(a-b) cz^2=by^2(a-b)/(c-a)=ax^2(a-b)/(b-c), by^2/(c-a)=ax^2/(b-c) or (C) by^2/(ax^2)=(c-a)/(b-c) The relative sizes of a, b and c can be: (1) a0 because b0 implies ac>0 and a>0 because c>0 (1C)<0 implies ab<0 and b<0 so ba: INCONSISTENT (2A)<0 implies bc<0 and b>0, c<0 because b>c (2B)<0 implies ac<0 and a>0 because c<0 (2C)>0 implies ab>0 and b>0 but c<0 and a>0 making ca: INCONSISTENT (3A)>0 implies bc>0 (3B)<0 implies ac<0 and a<0, c>0 and b>0 because a0 and so b>a, but b0 because b0 (4C)>0 implies ab>0, true, because a and b are both negative, but c0: INCONSISTENT (5A)>0 implies bc>0 (5B)<0 implies ac<0 and c<0, b<0, a>0 because a>c (5C)<0 implies ab<0, but b>a which cannot be if b<0: INCONSISTENT (6A)<0 implies bc<0 and c<0, b>0 because b>c (6B)>0 implies ac>0 and a<0, but b0 which cannot be: INCONSISTENT So there are no solutions because there is inconsistency throughout. (2) I am assuming this is a calculus problem. Assume a, b and c are constants and x, y and z are variables. adx/((b-c)y)=bdy/((c-a)x); b(b-c)ydy=a(c-a)xdx Integrating: b(b-c)y^2/2=a(c-a)x^2/2 (constant to be added later) a(c-a)x^2-b(b-c)y^2=k a constant. A similar result follows for the other two pairs of variables (x,z and y,z) by separation of variables and using the other two pairs of equations: a(a-b)x^2-c(b-c)z^2=p and b(a-b)y^2-c(c-a)z^2=q

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0 I think gaussian reduction is similar to back substitution, but with all the equation shaving different variables... just not sure how to do it. You create a matrix using the constants in the equations. Below, you will see approximately what the matrix would look like. Unfortunately, it is impossible to get it to display properly on this page. ┌                       ┐ │ 4   0   1  |   3  │ │ 2  -1  0   |  2   │ │ 0   3   2  |   0   │ └                        ┘ The idea is to perform the same math procedures on these matrix rows that you would perform on the full equations. We want the first row to be  1 0 0 | a, meaning that whatever value appears as the a entry is the value of x. The second row has to be  0 1 0 | b,  and the third row has to be  0 0 1 | c. Multiply row 2 by 2. ┌                      ┐ │ 4   0   1  |   3  │ │ 4  -2   0  |  4   │ │ 0   3   2  |   0   │ └                       ┘ Now subtract row 2 from row 1, and replace row 2 with the result. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  2   1   |  -1   │ │ 0   3   2  |   0   │ └                       ┘ Multiply row 2 by 2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  4    2  |  -2   │ │ 0   3   2  |   0   │ └                       ┘ Subtract row 3 from row 2, replacing row 2. ┌                        ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   3   2  |   0   │ └                        ┘ Multiply row 2 by 3 and subtract row 3, replacing row 3. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0  -2  |   -6  │ └                       ┘ Divide row 3 by -2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0   1  |   3   │ └                       ┘ Subtract row 3 from row 1, replacing row 1. ┌                      ┐ │ 4   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0  1  |   3   │ └                      ┘ Divide row 1 by 4. ┌                      ┐ │ 1   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0   1  |   3  │ └                      ┘ Row 1 shows that x = 0 Row 2 shows that y = -2 Row 3 shows that z = 3 Plug those values into the original equations to check the answer. 4x + z = 3 4(0) + 3 = 3 0 + 3 = 3 3 = 3 2x – y = 2 2(0) – (-2) = 2 0 + 2 = 2 2 = 2 3y + 2z = 0 3(-2) + 2(3) = 0 -6 + 6 = 0 0 = 0