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FinnZ [79.3K]
3 years ago
15

X^3+6x^2-17x+2-x^3-x^2-11x+36

Mathematics
1 answer:
Rudiy273 years ago
4 0
<span>X^3+6x^2-17x+2-x^3-x^2-11x+36 = </span>5x^2 - 28x + 38
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Solve for x:a(a²+b²)x²+b²x-a​
m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a  

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b²  = (b² + 2a²)²

2. Solve for x

\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}

\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}

5 0
3 years ago
**PLEASE HELP ILL MARK BRAINLIEST** Which of the following correctly represents y = 10x + 15
Vladimir [108]

Answer:

D

Step-by-step explanation:

Neither the table and the graph is correct.

7 0
3 years ago
Parallelogram JKLM has vertices J(-1, 6), K(0, 9), L(6, −3), and M(3, −3). What is the coordinates of the image if the parallelo
Digiron [165]

Answer: (\dfrac13,-2),\ (0,-3),\ (-2,1),\ (-1,1) .

Step-by-step explanation:

Transformation rule for dilation:

(x,y)\to(kx,ky) , where k = scale factor

Given : Scale factor = -\dfrac13

Parallelogram JKLM has vertices J(-1, 6), K(0, 9), L(6, −3), and M(3, −3)

Vertices after dilation:

(-1,6)\to(-1\times\dfrac{-1}{3},6\times\dfrac{-1}{3})=(\dfrac13,-2)

(0,9)\times(0\times-\dfrac13,9\times-\dfrac13)=(0,-3)

(6,-3)\to(6\times-\dfrac13,\ -3\times-\dfrac13)=(-2,1)

(3,-3)\to (3\times-\dfrac13,\ -3\times-\dfrac13)=(-1,1)

Hence, the coordinates of the image if the parallelogram = (\dfrac13,-2),\ (0,-3),\ (-2,1),\ (-1,1) .

8 0
3 years ago
100.6 + 296.5 using<br> mental math
ruslelena [56]

Answer:397.1....i think

Step-by-step explanation:

6 0
3 years ago
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Answer:

271 boxes

Step-by-step explanation:

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