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Ede4ka [16]
2 years ago
15

The sum of two polynomials is –yz2 – 3z2 – 4y 4. if one of the polynomials is y – 4yz2 – 3, what is the other polynomial? –2yz2

– 4y 7 –2yz2 – 3y 1 –5yz2 3z2 – 3y 1 3yz2 – 3z2 – 5y 7
Mathematics
1 answer:
kenny6666 [7]2 years ago
6 0

Polynomial are expression that have a leading degree of 3. The other polynomial function will be 3yz^2 – 3z^2 – 4y^4 - y + 3

<h3>Difference of polynomials</h3>

Polynomial are expression that have a leading degree of 3. Given the following expression

Sum = –yz^2 – 3z^2 – 4y^4.

One of the polynomial = y – 4yz^2 – 3

Required

The other polynomial

Let the unknown polynomial be "a" such that:

a =  –yz^2 – 3z^2 – 4y^4 - (y – 4yz^2 – 3)

Expand

a =  –yz^2 – 3z^2 – 4y^4 - y + 4yz^2 + 3

a = 3yz^2 – 3z^2 – 4y^4 - y + 3

Hence the other polynomial function will be 3yz^2 – 3z^2 – 4y^4 - y + 3

Learn more on polynomials here: brainly.com/question/4142886

#SPJ4

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Answer:

<h3>            x = -9,  y = -13 </h3><h3>    or    x = 13,   y = 9</h3><h3>    or    x = -13,  y = -9</h3><h3>    or     x = 9,   y = 13</h3>

Step-by-step explanation:

x^2+y^2=250\\\\x^2-2xy+y^2+2xy=250\\\\(x-y)^2=250-2xy\\\\(x-y)^2=250-2\cdot117\\\\ (x-y)^2=16\\\\x-y=4\qquad\qquad\vee\qquad \qquad  x-y=-4\\\\x=4+y \qquad\qquad \vee\qquad\qquad x=-4+y\\\\(y+4)y=117\qquad\vee\qquad\quad (y-4)y=117\\\\y^2+4y-117=0\qquad\vee\qquad y^2-4y-117=0\\\\y=\dfrac{-4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\qquad\vee\qquad y=\dfrac{4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\\\\y=\dfrac{-4\pm\sqrt{16+468}}{2}\qquad\ \ \vee\qquad y=\dfrac{4\pm\sqrt{16+468}}{2}

y_1=\dfrac{-4-22}{2}\ ,\quad y_2=\dfrac{-4+22}{2}\ ,\quad y_3=\dfrac{4-22}{2}\ ,\quad y_4=\dfrac{4+22}{2}\\\\y_1=-13\ ,\qquad y_2=9\ ,\qquad\quad\qquad\ y_3=-9\ ,\qquad y_4=13\\\\x_{1,2}=4+y_{1,2}\qquad\qquad\qquad\qquad\qquad x_{3,4}=-4+y_{3,4}\\\\x_1=-9\ ,\qquad x_2=13\ ,\qquad\quad\qquad x_3=-13\ ,\qquad x_4=9

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10. Complete the table below based on the following function:f(x)=-6x-1
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<span>f(x)=-6x-1

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WARRIOR [948]
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Read 2 more answers
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sleet_krkn [62]

Answer:

\huge \red { \boxed{x =  -  \frac{49}{17}}}

Step-by-step explanation:

log(3x + 1) - log(2x + 5) = 1 \\  \\ log  \: \frac{3x + 1}{2x + 5}  = log \: 10..( \because \: log \: 10 = 1) \\  \\  \frac{3x + 1}{2x + 5}  = 10 \\ \\  3x + 1 = 20x + 50 \\ 3x - 20x = 50 - 1 \\  \\  - 17x = 49 \\  \\ x =  \frac{49}{ - 17}  \\  \\  \huge \purple{ \boxed{x =  -  \frac{49}{17}}}  \\

6 0
2 years ago
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lozanna [386]

Answer:

  B.  0

Step-by-step explanation:

If you subtract the right side expression and simplify, you get ...

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5 0
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