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Veseljchak [2.6K]
3 years ago
5

What is the answer? up or down?

Mathematics
2 answers:
dolphi86 [110]3 years ago
8 0

Coefficient of x i.e a is negative so parabola will open downwards

Graph attached

Marat540 [252]3 years ago
5 0

What direction will the parabola open?

\sf \longrightarrow \: y =  -  {1x}^{2}  + 3x + 9

  • <em>The</em><em> </em><em>parabola</em><em> </em><em>will</em><em> </em><em>open</em><em> </em><em>in</em><em> </em><em>Downward</em><em> </em><em>direction</em><em>.</em>

<h3><u>How</u><u>?</u><u>?</u></h3>

  • <u>Remember</u><u> </u><u>that</u><u> </u><u>when</u><u> </u><u>the</u><u> </u><u>coefficient</u><u> </u><u>of</u><u> </u><u>x</u><u>²</u><u> </u><u>is</u><u> </u><u>negative</u><u> </u><u>the</u><u> </u><u>parabola</u><u> </u><u>opens</u><u> </u><u>in</u><u> </u><u>downward</u><u> </u><u>direction</u><u>.</u>
  • <u>When</u><u> </u><u>the</u><u> </u><u>coefficient</u><u> </u><u>of</u><u> </u><u>x</u><u>²</u><u> </u><u>is</u><u> </u><u>positive</u><u> </u><u>the</u><u> </u><u>parabola</u><u> </u><u>opens</u><u> </u><u>in</u><u> </u><u>upward</u><u> </u><u>direction</u><u>.</u><u> </u>
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Ann wants to choose from two telephone plans. Plan A involves a fixed charge of $10 per month and call charges at $0.10 per minu
Kay [80]
Ann wants to choose from two telephone plans. Plan A involves a fixed charge of $10 per month and call charges at $0.10 per minute. Plan B involves a fixed charge of $15 per month and call charges at $0.08 per minute.

Plan A $10 + .10/minute

Plan B $15 + .08/minute

If 250 minutes are used:

Plan A: $10+$25=$35
Plan B: $15+$20=$35

If 400 minutes are used:

Plan A: $10+$40=$50
Plan B: $15+$32=$47

B is the correct answer. How to test it:

Plan A: $10+(.10*249 minutes)
$10+$24.9=$34.9

Plan B: $15+(.08*249 minutes)
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4 0
3 years ago
find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0​
Citrus2011 [14]

Answer:

Radius: r =\frac{\sqrt {21}}{6}

Center = (-\frac{3}{2}, -\frac{2}{3})

Step-by-step explanation:

Given

9x^2 + 9y^2 + 27x + 12y + 19 = 0

Solving (a): The radius of the circle

First, we express the equation as:

(x - h)^2 + (y - k)^2 = r^2

Where

r = radius

(h,k) =center

So, we have:

9x^2 + 9y^2 + 27x + 12y + 19 = 0

Divide through by 9

x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0

Rewrite as:

x^2  + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}

Group the expression into 2

[x^2  + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}

[x^2  + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}

Next, we complete the square on each group.

For [x^2  + 3x]

1: Divide the coefficient\ of\ x\ by\ 2

2: Take the square\ of\ the\ division

3: Add this square\ to\ both\ sides\ of\ the\ equation.

So, we have:

[x^2  + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}

[x^2  + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2

Factorize

[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2

Apply the same to y

[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}

Add the fractions

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}

[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}

Recall that:

(x - h)^2 + (y - k)^2 = r^2

By comparison:

r^2 =\frac{7}{12}

Take square roots of both sides

r =\sqrt{\frac{7}{12}}

Split

r =\frac{\sqrt 7}{\sqrt 12}

Rationalize

r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}

r =\frac{\sqrt {84}}{12}

r =\frac{\sqrt {4*21}}{12}

r =\frac{2\sqrt {21}}{12}

r =\frac{\sqrt {21}}{6}

Solving (b): The center

Recall that:

(x - h)^2 + (y - k)^2 = r^2

Where

r = radius

(h,k) =center

From:

[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}

-h = \frac{3}{2} and -k = \frac{2}{3}

Solve for h and k

h = -\frac{3}{2} and k = -\frac{2}{3}

Hence, the center is:

Center = (-\frac{3}{2}, -\frac{2}{3})

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