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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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If h(x)=1-2/3x, find h(-6).
Alex

Answer:

  5

Step-by-step explanation:

Put -6 where x is and do the arithmetic.

  h(x)=1-\dfrac{2}{3}x\\\\h(-6)=1-\dfrac{2}{3}(-6)=1-\dfrac{-12}{3}=1-(-4)=1+4\\\\\boxed{h(-6)=5}

3 0
3 years ago
0.6 litres in the ratio 7:5
Sedaia [141]
Can you Elaborate your question?
5 0
2 years ago
Find the 7th term of the arithmetic sequence 5x + 9, 8x + 5, 11x + 1,...
ohaa [14]

Answer:

<h2>           \bold{a_7=33x-15}</h2>

Step-by-step explanation:

In arithmetic sequence:  a₂ - a₁ = a₃ - a₂     {common difference}

a_1=5x+9\\a_2=8x+5\\a_3=11x+1\\\\\\d=8x+5-(5x+9)=8x+5-5x-9=3x-4\\\\a_n=a_1+d(n-1)\\\\a_7=5x+9+(3x-4)\cdot6\\\\a_7=5x+9+18x-24\\\\a_7=33x-15

4 0
3 years ago
Use substitution to solve each system of equations.
uysha [10]

answers are in the pictures

7 0
2 years ago
K is the midpoint of PQ, P has
Pachacha [2.7K]

Answer:

Coordinates of Q (x_2,y_2) \:are\: \mathbf{(7,16)}

Option D is correct option.

Step-by-step explanation:

We are given:

K is the midpoint of PQ

Coordinates of P = (-9,-4)

Coordinates of K = (-1,6)

We need to find coordinates of Q  (x_2,y_2)

We will use the formula of midpoint: Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

We are given midpoint K and x_1,y_1 the coordinates of P we need to find x_2,y_2 the coordinates of Q.

Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\(-1,6)=(\frac{-9+x_2}{2},\frac{-4+y_2}{2})\\

Now, we can write

-1=\frac{-9+x_2}{2}, 6=\frac{-4+y_2}{2}\\Simplifying:\\-2=-9+x_2\:,\: 12=-4+y_2\\-2+9=x_2\:,\: 12+4=+y_2\\x_2=7\:,\:y_2=16

So, we get coordinates of Q (x_2,y_2) \:are\: \mathbf{(7,16)}

Option D is correct option.

6 0
3 years ago
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