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mina [271]
3 years ago
15

Help please I need it know

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
8 0
On 15 angle 1 and 2 would be 150 degrees i think, i could be wrong.
On 16 angle 1 would be 140 degrees and angle 2 would be 40 degrees. hope this helps
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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
I'm really confused about this question about Solving Systems by Elimination
velikii [3]

Answer:

x=10

y=2

Step-by-step explanation:

1)add the equations and find what x is.

x-y=8

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—————

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— —

2 2

x=10

3)fill in x for the second equation.

x+y=12

since x=10,replace x with 10.

10+y=12.

subtract 19 from both sides.

10+y=12

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—————-

y=2

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amm1812
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19 x 9 short division
Colt1911 [192]

Answer:

170

Step-by-step explanation:

your welcome np 8 hope you appreciate it

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