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can you help me solve this ASAP [3](8*4)•6÷(4-2)

[3](8×4)•6÷(4-2)

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I NEED help! Asap can someone please please help me solve 6 ...


I NEED help! Asap can someone please please help me solve 6 and 7!
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Can You Solve This? - YouTube


Feb 23, 2014 · Can You Solve This? Veritasium. Loading... ... History Help About; Press; Copyright; Creators; Advertise; Developers +YouTube; Terms; Privacy; Policy ...

Solve : 2x -1>3 Please an someone help me solve this ASAP ...


Solve : 2x -1>3 Please an someone help me solve this ASAP 2. Ask for details ; Follow; ... NEED HELP ASAP PLEASE!! Part A: Which is the correct net for the prism?
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URGENT: can someone help me solve this ASAP The... - OpenStudy


URGENT: can someone help me solve this ASAP The cube root of 4 times the ... can someone help me solve this ASAP The cube root of 4 times the sixth root of 8 ...
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can you solve the quadratic function? y = 2x^2 + 3 - jiskha.com


... exponential function or none of the above. x^2+ 6 = -8 x quadratic equations can you solve ... need help like ASAP! You can ... 1 6.8,2,27.2,3 108.8, 4 435 ...
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Please help ASAP - FreeMathHelp.com


Please help ASAP Hello ... 5 2 6 4 7 5 8 6 9 3 The question is find the mean of these shoe sizes: choose one: a) ... Solve for [tex]x. ...
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need help ASAP with 3+ sq rootx^2-8x=0 - FreeMathHelp.com


Hi-- can anyone help me solve this?? 3+ sq rootx^2-8x=0 thanks! Help; Remember Me? What's New? Forum; FAQ ... need help ASAP with 3+ sq rootx^2-8x=0
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solve the system of equations y=2x^2 -3 and y=3x-1

kerv 1: y=2x^2 -3 kerv 2: y=3x-1 hit point...2x^2-3=3x-1 or 2x^2 -3 -3x +1=0 2x^2 -3x -2=0 (x-2)(x+0.5)=0
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Help needed in understanding the process to solve a trig problem such as [ 2sin^2x-3x= 0 ]

  Remember the relation between sine and cosine? sine^2+cosine^2=1, so sin^2X=1-cos^2X. Substitute for sin^2X: 2(1-cos^2X)-3cosX=0; 2-2cos^2X-3cosX=0. We now have a quadratic equation. If it helps, let y= cosX, then: 2-2y^2-3y=0 or 2y^2+3y-2=0, which factorises: (2y-1)(y+2)=0, and y=1/2 or -2 which means cosX=1/2 or -2. But we know that cosX can't be bigger than 1 or less than -1, so we can safely reject cosX=-2, and solve cosX=1/2. This is one of those familiar angles, X=60 degrees or (pi)/3 radians. Now you need to remember which quadrants cosine is positive in. ASTC (All Silver Tea Cups: All trigs, sin, tan, cos) for Q1 to 4. So cosine is positive in Q1 and Q4. The solution therefore is (pi)/3 and 2(pi)-(pi)/3=5(pi)/3 or 60 and 360-60=300 degrees.
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Help understanding the quadratic equation

if the quadratic eqn is ax^2 + bx+c = 0 you can use the formula -a +/- sqrt(b^2-4ac) / (2a) in your case, all can be factored 1. (x-5)(x+1) = 0 2. (z-5)(z+3) = 0 3. (y-3)(y+1) = 0 4. (r+5)(r-1) = 0
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solve x²+7x+12= 0

solve x²+7x+12= 0 I need to know how to work it out - please help!!!!! x² + 7x + 12 = 0 We need to find factors of 12 that add up to 7. They must be 3 and 4 (x + 3)(x + 4) = 0 One, or both, of those factors on the left must equal zero. (Multiplying by zero produces a zero result.) Set each one to zero and solve for x. (x + 3) = 0 x = -3 (x + 4) = 0 x = -4 Check both answers using the original equation. x² + 7x + 12 = 0 -3² + 7(-3) + 12 = 0 9 - 21 + 12 = 0 0 = 0 x² + 7x + 12 = 0 -4² + 7(-4) + 12 = 0 16 - 28 + 12 = 0 0 = 0 So, x = -3 and x = -4 are solutions. If you graph this equation, you will get a parabola that crosses the x axis at two points: (-3, 0) and (-4, 0)  
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solve y= 5x+2, x=6-3y. this is systems of equations :)

Problem: solve y= 5x+2, x=6-3y. this is systems of equations :) please help me l0l 1) y = 5x + 2 2) x = 6 - 3y Substitute the value of y, from equation 1, into equation 2. x = 6 - 3y x = 6 - 3(5x + 2) x = 6 - 15x - 6 x + 15x = 6 - 15x - 6 + 15x 16x = 6 - 6 16x = 0 x = 0 Use that to solve for y. y = 5x + 2 y = 5(0) + 2 y = 0 + 2 y = 2 Check. x = 6 - 3y 0 = 6 - 3(2) 0 = 6 - 6 0 = 0 Answer: x = 0, y = 2  
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Guys can you help me solve these following equations

This should help: Simplifying 3p + -1 = 5(p + -1) + -2(7 + -2p) Reorder the terms: -1 + 3p = 5(p + -1) + -2(7 + -2p) Reorder the terms: -1 + 3p = 5(-1 + p) + -2(7 + -2p) -1 + 3p = (-1 * 5 + p * 5) + -2(7 + -2p) -1 + 3p = (-5 + 5p) + -2(7 + -2p) -1 + 3p = -5 + 5p + (7 * -2 + -2p * -2) -1 + 3p = -5 + 5p + (-14 + 4p) Reorder the terms: -1 + 3p = -5 + -14 + 5p + 4p Combine like terms: -5 + -14 = -19 -1 + 3p = -19 + 5p + 4p Combine like terms: 5p + 4p = 9p -1 + 3p = -19 + 9p Solving -1 + 3p = -19 + 9p Solving for variable 'p'. Move all terms containing p to the left, all other terms to the right. Add '-9p' to each side of the equation. -1 + 3p + -9p = -19 + 9p + -9p Combine like terms: 3p + -9p = -6p -1 + -6p = -19 + 9p + -9p Combine like terms: 9p + -9p = 0 -1 + -6p = -19 + 0 -1 + -6p = -19 Add '1' to each side of the equation. -1 + 1 + -6p = -19 + 1 Combine like terms: -1 + 1 = 0 0 + -6p = -19 + 1 -6p = -19 + 1 Combine like terms: -19 + 1 = -18 -6p = -18 Divide each side by '-6'. p = 3 Simplifying p = 3
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Whats 3x-2 over 9 equals 25 over 3x-2?

I will give you an example for this question. This might be able to help you. Simplifying 3x2 + 25x = 18 Reorder the terms: 25x + 3x2 = 18 Solving 25x + 3x2 = 18 Solving for variable 'x'. Reorder the terms: -18 + 25x + 3x2 = 18 + -18 Combine like terms: 18 + -18 = 0 -18 + 25x + 3x2 = 0 Factor a trinomial. (-9 + -1x)(2 + -3x) = 0 Subproblem 1 Set the factor '(-9 + -1x)' equal to zero and attempt to solve: Simplifying -9 + -1x = 0 Solving -9 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '9' to each side of the equation. -9 + 9 + -1x = 0 + 9 Combine like terms: -9 + 9 = 0 0 + -1x = 0 + 9 -1x = 0 + 9 Combine like terms: 0 + 9 = 9 -1x = 9 Divide each side by '-1'. x = -9 Simplifying x = -9 Subproblem 2 Set the factor '(2 + -3x)' equal to zero and attempt to solve: Simplifying 2 + -3x = 0 Solving 2 + -3x = 0 Move all terms containing x to the left, all other terms to the right. Add '-2' to each side of the equation. 2 + -2 + -3x = 0 + -2 Combine like terms: 2 + -2 = 0 0 + -3x = 0 + -2 -3x = 0 + -2 Combine like terms: 0 + -2 = -2 -3x = -2 Divide each side by '-3'. x = 0.6666666667 Simplifying x = 0.6666666667 Solution x = {-9, 0.6666666667}
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how to solve area under the curve x = y^2 - 2y , y=0 , x = -1

It's best to sketch the curve first. This helps to picture how to find the area under the curve. It's a sideways parabola. In a way, you can swap the x and y variables so that the graph looks like y=x^2-2x but the axes are interchanged to produce x=y^2-2y. When x=0, y^2-2y=0=y(y-2) so y=0 (origin) and 2 are the y intercepts. When y=0, x=0 (origin). The vertex is when 2y-2 (the derivative with respect to y) is zero, so y=1 and x=-1. The vertex is at (-1,1). The second derivative is 2, so the turning point is a minimum. The graph is U-shaped lying on its side with arms pointing to the right (positive). The parabola has its vertex in quadrant 2 and its arms cross the y axis at 0 and 2. x=y^2-2y, but what is y in terms of x? completing the square we have y^2-2y+1=1+x, (y-1)^2=1+x and y=1+sqrt(1+x). There are generally two values of y for the same value of x. The graph illustrates this, because the parabola is lying on its side so the x value picks up two values of y except at the vertex. We need the lower part of the graph only, given by y=1-sqrt(1+x). If on the graph we paint the inside of the parabola blue and the outside red, this will help us to identify the areas. We want, I think, the red area between the vertex and the x axis and, I assume, up to the y axis, because the curve dips below the x axis when x is positive. Imagine thin rectangles width dx lying vertically with a length equal to y=1-sqrt(1+x). The length of the rectangles for y=1+sqrt(x) include blue and red areas. The area of each rectangle is ydx and the area under the graph is the sum total of the areas of the thin rectangles. When the rectangles are infinitely thin this sum total is the integral of ydx. If we wanted, for example, just the blue area we would need to subtract the red area from the blue-and-red area. This is why it's important to sketch a graph and understand how to apply the mathematics. So our thin rectangles start at x=-1, when y=1, so this rectangle has a width dx and a length of 1; and ends with a rectangle of zero height when x=0. If we integrate between the x limits of -1 and 0 we will get the red area we need: integral((1-sqrt(1+x))dx)=integral((1-(1+x)^1/2)dx)=[x-2(1+x)^(3/2)/3](-1 Read More: ...

solve by factoring 9x^2=-18x

9x^2 = -18x to solve you need to set it to 0 9x^2 + 18x = 0 9x(x + 2) = 0 9x = 0 or x + 2 = 0 x = 0 or x = -2 CHECK: 9* 0^2 + 18(0) = 0 yes true 9*(-2)^2 + 18(-2) = 0 9*4 - 36 = 0 36 - 36 = 0 yes true therefore x = 0, -2          
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help! Find all x-intercepts and y-intercepts, if no intercepts exist say so

Problem: help! Find all x-intercepts and y-intercepts, if no intercepts exist say so f(x)=5x-8     f(x)=3x ,  f(x)=5 i dont get how to solve these and find the intercepts.... In place of f(x), substitute y. To find the y-intercept, set x to zero. To find the x-intercept, set y to zero. f(x) = 5x - 8 y = 5x - 8 y = 5(0) - 8 y = -8, the y-intercept y = 5x - 8 0 = 5x - 8 5x = 8 x = 8/5, the x-intercept f(x) = 3x y = 3x y = 3(0) y = 0, the y-intercept y = 3x 0 = 3x 3x = 0 x = 0/3 = 0, the x-intercept f(x) = 5 y = 5, the y-intercept There is no x term, so there is no x-intercept This represents a horizontal line five units above the x-axis.
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