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bonufazy [111]
3 years ago
6

Which inequality does the graph below represent?

Mathematics
1 answer:
Dima020 [189]3 years ago
7 0

Answer:

A. y\le2x^2-8x+3

Step-by-step explanation:

The given parabola has vertex at (2,-5).

The equation of this parabola in vertex form is given by:

y=a(x-h)^2+k, where (h,k)=(2,-5) is the vertex  of the parabola.

We substitute the values to get:

y=a(x-2)^2-5

The graph passes through; (0,3).

3=a(0-2)^2-5

\implies 3+5=4a

\implies 8=4a

\implies a=2

Hence the equation of the parabola is

y=2(x-2)^2-5

We expand this to get:

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

y=2x^2-8x+3

Since the outward region was shaded, the corresponding inequality is

y\le2x^2-8x+3

The correct answer is A

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4 0
2 years ago
Given that tan^2 theta=3/8,what is the value of sec theta?
algol [13]

Answer:

The value of SecФ is  \pm \sqrt{\frac{11}{8}} .

Step-by-step explanation:

Given as for trigonometric function :

tan²Ф = \frac{3}{8}

Or, tanФ = \sqrt{\frac{3}{8} }

∵ tanФ = \frac{Perpendicular}{Base}

So,  \frac{Perpendicular}{Base} =  \sqrt{\frac{3}{8} }

So, Hypotenuse² = perpendicular² + base²

or, Hypotenuse² = ( \sqrt{3} )² + ( \sqrt{8} )²

Or,  Hypotenuse² = 3 + 8 = 11

Or,  Hypotenuse = ( \sqrt{11} )

Now SecФ = \frac{Hypotenuse}{Base}

or, SecФ = \frac{\sqrt{11}}{\sqrt{8}} = \sqrt{\frac{11}{8} }

<u>Second Method</u>

Sec²Ф - tan²Ф = 1

Or, Sec²Ф = 1 +  tan²Ф

or, Sec²Ф = 1 +  \frac{3}{8}

Or, Sec²Ф = \frac{11}{8}

Or,  SecФ = \pm \sqrt{\frac{11}{8}}

Hence The value of SecФ is  \pm \sqrt{\frac{11}{8}} . Answer

7 0
3 years ago
Read 2 more answers
Give right triangle PQR, which representa the value of sin (P)?
Komok [63]

sinR = \frac{QR}{PQ}

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4 0
3 years ago
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liubo4ka [24]

Answer:

X=-6

Step-by-step explanation:

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Step 2: Add 2x to both sides.

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Step 3: Subtract 4 from both sides.

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Answer:

x=−6

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