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Cerrena [4.2K]
3 years ago
14

PLZ HElPPPPPP MEEEEEEEEE !

Mathematics
1 answer:
Effectus [21]3 years ago
7 0
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Find three solutions of the equation y=9x-4?
Bas_tet [7]
<span>answer
A) (-5,-49),(-2,-22),(3,23)

</span>when x = - 5 ; y=9x-4 = 9(-5) -4  = -49
when x = - 2 ; y=9x-4 = 9(-2) -4  = -22
when x = 3 ; y=9x-4 = 9(3) -4  = 23
6 0
3 years ago
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I need you to answer with a, b, c, d
solong [7]

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

\begin{gathered} ax^2+bx+c=0 \\  \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}

For the given function:

f(x)=2x^2-10x-3x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}x=\frac{10\pm\sqrt[]{100+24}}{4}\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\  \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\  \end{gathered}\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\  \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\  \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Then, the zeros of the given quadratic function are:

\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\  \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Answer: Third option

8 0
1 year ago
Which of the following is equal to [(x2y3)-2/(x6y3x)2]3 ?
Alona [7]

Answer:

a=√x3y5 or a=−√x3y5

Step-by-step explanation:

4 0
2 years ago
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Indicate true or false as to whether the following equation is quadratic 2x-4x+1=0
saul85 [17]
The answer to the problem would be false
3 0
3 years ago
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How to solve (x+3)^2+7=29
tangare [24]
X^2 + 9 + 7 = 20
X^2 +16 = 20
X^2 = 20-16
X^2 = 4
Root 4 = 2
X=2
6 0
3 years ago
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