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Math
Calculus
Integral [-1,6] 12x^2–x^3–32x
1 answer
straightforward power integrals. What do you get?
∫12x^2 dx = 4x^3, etc.
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Can someone help me with this please.
Andrew factored the expression 20x^3-12x^2+8x as 4x(5x^2-12x^2+8x) but when melissa applied
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Andrew divided only the first term by 4x. Should have been 4x(5x^2 - 3x + 2)
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Multiply (x-5/4x+8)*(12x^2+32x+16)
A. (3x+2)/4(x-5) B. (x-5)(3x+2)/4 C. (x-5)(3x+3) D. (x-5)(12x+8) Please help me solve this
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I will assume that you meant (x-5)/(4x+8)*(12x^2+32x+16) that simplifies to (x-5)/4(x+2) *
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Can someone check my work and answer?
Evaluate the integral from -1 to 0 of (4x^6+2x)^3(12x^5+1)dx My work: let u=4x^6+2x
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OK. Let's see here... ∫(4x^6+2x)^3(12x^5+1)dx let u = 4x^6+2x du = 24x^5+2 = 2(12x^5+1) so, you
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Andrew factored the expression 20x^3-12x^2+8x as 4x (5x^2-12x^2+8x). But when Melissa applied the distributive law and
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sure, you can take 4x out but do it properly 4x (5 x^2 -3 x + 2)
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Andrew factored the expression 20x^3-12x^2+8x as 4x(5x^2-12x^2+8x). But when Melissa applied the distributive law and multiplied
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Andrew messed up, should have been = 4x(5x^2 - 3x + 2)
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Multiply. x-5/4x+8*(12x^2+32x+16)
A. (3x+2)/4(x-5) B. (x-5)(3x+2)/4 C. (x-5)(3x+2) D. (x-5)(12x+8)
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The answer is C. (x-5)(3x+2). To multiply, we first distribute the x-5 over the parentheses:
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2x^3 - 12x^2 - 32x
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factor? 2x*(x^2-6x-16) 2x(x+2)(x-8)
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X-5/4x+8 * (12x^2+32x+16)
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To simplify the expression: First, factor out the common factor from the first term: = (x - 5)/(4x +
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-12x^3+32x^4
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The given polynomial is: -12x^3 + 32x^4
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Write y−2=32(x+2) in slope-intercept form.(1 point)
Responses y=32x+5 y=32x+6 y=32x+32 y=32+5x
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The correct answer is y=32x+68. To convert the given equation to slope-intercept form, we need to
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(x-5)/(4x+8)*(12x^(2)+32x+16)
multiply
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12x^2-32x-12
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X-5/4x+8 * (12x^2+32x+16)
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