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# AC=3x+3, AB=-1+2x, BC=11 Find x

Please i need help can someone show me the steps of showing the work because i don't understand how to solve.

## Research, Knowledge and Information :

### SOLUTION: B is between A and C. AB = 2x - 3, BC = 3x + 1, AC ...

AC = AB + BC (7X-8) = (2X-3) + (3X+1) Substitute the values of X=3 and see that LHS = RHS of the equation. let us do that, 7*3-8 = 2*3-3 + 3*3+1 21-8 = 6-3+9+1

### Homework Help ...

... and BC=11 find X ... More. Spanish Economics Geography Vocabulary French Accounting ... AC= 3X + 3, AB= -1+2 X , and BC=11 find X – Umisays9. Umisays9 ...

### The Segment Addition Postulate Date Period

The Segment Addition Postulate Date_____ Period____ Find the ... x. 11) AC = x , AB = x, ... AB = x, BC = , AC = ...

### B is the midpoint of AC. If Ab= x +5 and BC = 2x-11, find the ...

If Ab= x +5 and BC = 2x-11, find the measure of Ab. - 3450428. 1. ... B is the midpoint of AC. If Ab= x +5 and BC = 2x-11, ... (x + 5) + (2x − 11) = 3x − 6 ...

### Point B is between A and C on segment AC. Use the given ...

Use the given information to write an equation in terms of x. ... Then find AB and BC. AB= 3x; BC= x; AC= 20 AB= 2x-5; BC= 6x; ... 2016-11-22T01:47:16-05:00.

## Suggested Questions And Answer :

### what are the critical numbers of the equation: 3x^4-8x^3+6x^2

f(x) = 3x^4 - 8x^3 + 6x^2         Find the derivative of the function f'(x) = 12x^3 - 24x^2 +12x       The critical values are the f' set = 0   12x^3 -24x^2 + 12x  = 0               12x(x^2 - 2x + 1) = 0               factor out 12x 12x (x - 1) (x - 1) = 0               factor the x^2 factor 12x = 0 , x - 1 = 0 x = 0 , x = 1 restating the f' f'(x) = 12x^3 - 24 x^2 + 12x   find the second derivative f"(x) = 36x^2 - 48x + 12 f"(0) = 12 positive   that means there is a min at x = 0. f"(1) = 36(1) - 48(1) + 12 = 36 -48 + 12 = 0  at x = 1 there is a invertion point.

### find and classify all the critical points of the function f(x)=2x^3+3x^2-12x+2

Question: find and classify all the critical points of the function f(x)=2x^3+3x^2-12x+2. f = 2x^3 + 3x^2 - 12x + 2 differentiating, f' = 6x^2 + 6x - 12 Critical points, or stationary points, are those points on the curve of f(x) where the slope is zero, i.e. the derivative of f(x) is zero. Setting f'(x) = 0, 6x^2 + 6x - 12 = 0 x^2 + x - 2 = 0 (x - 1)(x + 2) = 0 x = 1, x = -2 2nd derivative, f''(x) f''(x) = 12x + 6 at x = 1 f''(1) = 12 + 6 = 18 > 0 Since f'' > 0,then the turning point is a minimum. at x = -2 f''(-2) = -24 + 6 = -18 < 0 Since f'' < 0,then the turning point is a maximum. f(x) = 2x^3 + 3x^2 - 12x + 2 at x = 1, f(1) = 2(1) + 3(1) - 12(1) + 2 = 2 + 3 - 12 + 2 = -5 at x = -2, f(-2) = 2(-8) + 3(4) - 12(-2) + 2 = -16 + 12 + 24 + 2 = 22 The critical points are: (1, -5) min, (-2, 22) max

### Let g(x) = 2x-3 and h(x) = 2x^2 + 4x -8. Find f(x) such that fog (composition) = h

Let g(x) = 2x-3 and h(x) = 2x^2 + 4x -8. Find f(x) such that fog (composition) = h To find f() s.t. fog = h g^2 = 4x^2 – 12x + 9 Now, h = 2x^2 + 4x + 8 h = (4x^2 + 8x + 16)/2 h = (4x^2 – 12x + 20x + 9 + 7)/2 h = (4x^2 – 12x + 9)/2 + (20x + 7)/2 h = g^2/2 + 10x + 3 ½ h = g^2/2 + 10x – 15 + 15 + 3 ½ h = g^2/2 + 5(2x – 3) + 18 ½ h = g^2/2 + 5g + 18 ½ And, fog = h, i.e. f(g) = g^2/2 + 5g + 18 ½ f(x) = (x^2 + 10x + 37)/2

### Use addition or Substitution to find the value of x for this set of equations

Problem: Use addition or Substitution to find the value of x for this set of equations 12x-5y=30 y=2x-6 1) 12x - 5y = 30 2) y = 2x - 6 Might as well use substitution. 12x - 5y = 30 12x - 5(2x - 6) = 30 12x - 10x + 30 = 30 2x + 30 = 30 2x + 30 - 30 = 30 - 30 2x = 0 x = 0

### how do you find the inverse of this function? f(x) = 12x + 50

y=12x+50 invers... 12x=y-50....x=(y-50)/12 or (y/12)-4.1666666666666666666666...

### Pete challenges his friend Jill to find two consecutive odd integers that have the following relationship.

Two consecutive odd integers can be represented by 2x-1 and 2x+1. Their product is 4x^2-1 and their sum is 4x. So, 4x^2-1=3(4x+6)=12x+18. This is the quadratic: 4x^2-12x-19=0. This has no rational solutions. Or, reading the question slightly differently: 4x^2-1=12x+6, so 4x^2-12x-7=0. In the second case, we can factorise: (2x-7)(2x+1)=0, so x=3.5 or -0.5. The two integers are: 6 and 8, or -2 and 0. None of these are odd, so it isn't possible to find two consecutive odd integers. Why did I think there were two interpretations of the question? "3 times the sum of the integers plus 6" is ambiguous:  is it 3(sum+6) or 3*sum+6? Judging by the result, it was the latter.

### How do i find what "x" equals?

Problem: How do i find what "x" equals? 9+12x=81 AIG work for 6th garde. 9 + 12x = 81 9 + 12x - 9 = 81 - 9 12x = 72 12x/12 = 72/12 x = 6

### given px-13+2x+y=0.py-13+x+2y=0 If Tc=x+y find the price and output for each good which will maximize profit shown using the second order condition that profit have a maximum at this point

Price of input, Px=13-2x-y; price of output, Py=13-x-2y; total cost, Tc=x+y. Revenue comes from the sale of the output=y*Py. Variable cost comes from production of the input=x*Px. Let the fixed cost be F, profit P=(revenue)-(total cost of production)= yPy-xPx-F=13y-xy-2y^2-13x+2x^2+xy-F= 13(y-x)-2(y^2-x^2)-F=13(y-x)-2(y-x)(y+x)-F=(y-x)(13-2(y+x))-F, and total cost is cost of production xPx+F=Tc=x+y (given). So 13x-2x^2-xy+F=x+y. From this, 12x-2x^2+F=y(1+x) and y=(2x(6-x)+F)/(1+x). This relates output and input quantities. So, dividing by 1+x: y=2(7-x)+(F-14)/(1+x). Since y>0, F>2x(x-6) and x>6 because F cannot be negative. Px=13-2x-(2x(6-x)+F)/(1+x)=(13+13x-2x-2x^2-12x+2x^2-F)/(1+x)=(13-F-x)/(1+x). Px>0, so F<13-x, and x<13. Now we can see that 62x(x-6), so 00. Therefore, 13-2F-12x+3x^2>0 because Py>0. x^2-4x+(13-2F)/4>0; x^2-4x+4+(13-2F)/4-4>0; (x-2)^2>(3+2F)/4. This implies x>2±sqrt(3/4+F/2)>0 and since we know 6 Read More: ...

### solve the system: 5x-10y+4z=-73; -x+2y-3z=19; 4x-3y=5z=-42

solve the system: 5x-10y+4z=-73; -x+2y-3z=19; 4x-3y=5z=-42 1)  5x - 10y + 4z = -73 2)  -x +  2y - 3z =  19 3)  4x -  3y + 5z = -42 Equation three had "=5z" so I changed the equal sign to a plus sign. It happens a lot; the shift key doesn't register, so what was supposed to be "+" comes out "=". We're going to eliminate z so we can work on finding x and y. We need the coefficients of z to be equal (disregarding the sign). Equation one; multiply by 3: 3 * (5x - 10y + 4z) = -73 * 3 4)  15x - 30y + 12z = -219 Equation two; multiply by 4: 4 * (-x +  2y - 3z) =  19 * 4 5)  -4x + 8y - 12z = 76 Add equation five to equation four, eliminating z.   15x - 30y + 12z = -219 +(-4x +  8y - 12z =   76) -----------------------------------   11x - 22y         = -143 6)  11x - 22y = -143 Follow the same procedure with equations two and three. Equation two: multiply by 5 this time: 5 * (-x +  2y - 3z) =  19 * 5 7)  -5x + 10y - 15z = 95 Equation three: multiply by 3: 3 * (4x -  3y + 5z) = -42 * 3 8)  12x - 9y + 15z = -126 Add equation eight to equation seven, eliminating z.    -5x + 10y - 15z =    95 +(12x -  9y + 15z = -126) ------------------------------------     7x +    y         =  -31 9)  7x + y = -31 Multiply equation nine by 22; then we will add equation six. 22 * (7x + y) = -31 * 22  154x + 22y = -682 +(11x - 22y = -143) ---------------------------  165x         = -825 x = -5  <<<<<<<<<<<<<<<<<<<<<<<<<< Substitue that into equations six and nine to find y. They will serve to verify the calculations if you do it with two equations. 11x - 22y = -143 11(-5) - 22y = -143 -55 - 22y = -143 -22y = -143 + 55 -22y = -88 y = 4  <<<<<<<<<<<<<<<<<<<<<<<<<< 7x + y = -31 7(-5) + y = -31 -35 + y = -31 y = -31 + 35 y = 4     same value for y Now, we can go back to the original equations, plug in both x and y, and solve for z. To check and verify, we will use all three of the original equations. One: 5x - 10y + 4z = -73 5(-5) - 10(4) + 4z = -73 -25 - 40 + 4z = -73 -65 + 4z = -73 4z = -73 + 65 4z = -8 z = -2  <<<<<<<<<<<<<<<<<<<<<<<<<< Two: -x +  2y - 3z =  19 -(-5) +  2(4) - 3z =  19 5 + 8 - 3z = 19 13 - 3z = 19 -3z = 19 - 13 -3z = 6 z = -2     same answer for z Three: 4x -  3y + 5z = -42 4(-5) -  3(4) + 5z = -42 -20 - 12 + 5z = -42 -32 + 5z = -42 5z = -42 + 32 5z = -10 z = -2    again, verified x = -5, y = 4, z = -2

### f(x)=6x^2 - 5x + 4 find the slope between x=-2 and x=a

The slope varies according to the derivative, f'(x)=12x-5. At f'(-2) the slope is 12(-2)-5=-29. At f'(-3) it is 12(-3)-5=-41. So the slope varies from point to point. If you need to find the slope between two points, the best you can get is the average, so in this case we would have the average of -29 and -41=-70/2=-35. Negative means a backwards slope (\). Using h as the difference between two values of x we have: f(x)=6x^2-5x+4 and f(x+h)=6(x+h)^2-5(x+h)+4. If we calculate f(x+h)-f(x) we get: 6((x+h)^2-x^2)-5h=6(x+h-x)(x+h+x)-5h=6h(2x+h)-5h=12hx+6h^2-5h. However, h is supposed to be very small so that h^2 can be ignored. If h is 1, for example, it's too big to be ignored so that when we divide through by h, we get 12x+6h-5 instead of 12x-5, the derivative as shown above. Instead of -29 (for x=-2) we would have -29+6h. If h=-1, this becomes -35, which is the average slope.