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3 years ago

Whats the vertex of a parabola? Round to the nearest hundredth.

Mathematics

2 answers:

olya-2409 [2.1K]3 years ago

4 0

Answer:

To get the x-intercept we need the equation of the parabola;

the vertex form of the equation is given by:

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

where:

(2,13) is the vertex:

Therefore;

y=a(x-2)^2+13

when x=0, y=5

Therefore the value of a will be:

5=a(0-2)^2+13

5-13=4a

4a=-8

a=-2

therefore the equation of the parabola will be:

y=-2(x-2)^2+13

y=-2x^2+8x+5

the x-intercept will be:

x=2-sqrt(7.5)

x=2+sqrt(7.5)

Step-by-step explanation:

professor190 [17]3 years ago

4 0

Answer:

Step-by-step explanation:

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ioda

Answer with explanation:

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2 years ago

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maria [59]

Answer:

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Step-by-step explanation:

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2 years ago

If sin A = 3/8, find the value of cosec A - sec A.

alexira [117]

Answer:

$\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

Step by step explanation:

$\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies 1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\$

$\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}$

$\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$