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qaws [65]
3 years ago
14

Which equation represents a parabola that opens upward has a minimum at x=3 and has a line of symmetry at x=3

Mathematics
2 answers:
Advocard [28]3 years ago
5 0

Answer:

A. y = x^2 -6x + 13

Step-by-step explanation:

umka2103 [35]3 years ago
4 0

Answer:

y=x^2-6x+5

Step-by-step explanation:

Let us consider the equation y=x^2-6x+5

For a quadratic equation in a standard form, y=ax^2+bx+c, the axis of symmetry is the vertical line x = \frac{-b}{2a}.

Here in this case we have, a=1, b=-6 , c =5

Putting the values we get,

x = \frac{-(-6)}{2\times 1} = \frac{6}{2} =3

We can see that the axis of symmetry is x=3 and the graph is giving minimum at x=3.

Therefore, the required equation is y=x^2-6x+5. Refer the image attached.


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Answer:

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⠀

Step-by-step explanation:

The triangle having two sides equal along with one different side is called an isosceles triangle.

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  • So, Let us assume the other sides (two equal sides) of the isosceles triangle as x. As the different side is already given in the question.

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We know that,

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So, According to the question :

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{\longrightarrow \it\qquad { \ { {  38 \: }^{ \circ}   + x + x  =  180{}^{ \circ} }}}

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{\longrightarrow \it\qquad { \ { {  38 \: }^{ \circ}   + 2x   =  180{}^{ \circ} }}}

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{\longrightarrow \it\qquad { \ {   2x   =  180{}^{ \circ} - {  38 \: }^{ \circ}}}}

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{\longrightarrow \it\qquad { \ {   2x   =  142{}^{ \circ}  }}}

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{\longrightarrow \it\qquad { \ {   x   =    \dfrac{142{}^{ \circ}}{2}   }}}

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{\longrightarrow \it\qquad { \pmb {   x   =  71^{ \circ}  }}}

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Therefore,

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