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Mama L [17]
3 years ago
11

J(x) = 2x²+8x-5 What is this in vertex form

Mathematics
1 answer:
pav-90 [236]3 years ago
5 0

Answer:

The vertex form parabola   y = 2( x+4)² -37    

Step-by-step explanation:

<u>Step(i):-</u>

Given parabola equation  j(x) = 2x² + 8x -5

Let    y = 2x² + 8x -5

⇒   y = 2(x² + 2(4x)+(4)²-(4)²) -5

By using  (a + b)² = a² +2ab +b²

       y = 2(x+4)²- 32 -5    

        y = 2 ( x-(-4))² -37      

<u><em>Step(ii):-</em></u>

The vertex form parabola   y = a( x-h)² +k

The vertex form parabola   y = 2(x+4)² -37    

   

                   

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The area of a rectangular bedroom is 240 square feet and the length is 16 feet what is the width
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2.5mph

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During halftime of a basketball ​game, a sling shot launches​ T-shirts at the crowd. A​ T-shirt is launched from a height of 4 f
Sonbull [250]

Answer:

(a) The time the T-shirt takes to maximum height is 2 seconds

(b) The maximum height is 68 ft

(c) The range of the function that models the height of the T-shirt over time given above is 4 + 64\cdot t - 16 \cdot  t^{2}

Step-by-step explanation:

Here, we note that the general equation representing the height of the T-shirt as a function of time is

h = h_1 + u\cdot t - \frac{1}{2} \cdot g  \cdot  t^{2}

Where:

h = Height reached by T-shirt

t = Time of flight

u = Initial velocity = 64 ft/s

g = Acceleration due to gravity (negative because upward against gravity) = 32 ft/s²

h₁ = Initial height of T-shirt = 4 ft

(a) The maximum height can be found from the time to maximum height given as

v = u - gt

Where:

u = Initial velocity = 64 ft/s

v = Final upward velocity at maximum height = 0 m/s

g = 32 ft/s²

Therefore,

0 = 64 - 32·t

32·t = 64 and

t = 64/32 = 2 seconds

(b) Therefore, maximum height is then

h = 4 + 64\times 2 - \frac{1}{2} \times 32  \times  2^{2}

∴ h = 68 ft

The T-shirt is then caught 41 ft above the court on its way down

(c) The range of the function that models the height of the T-shirt over time given above is derived as

h = h_1 + u\cdot t - \frac{1}{2} \cdot g  \cdot  t^{2}

With u = 64 ft/s

g = 32 ft/s² and

h₁ = 4 ft

The equation becomes

h =4 + 64\cdot t - \frac{1}{2} \times 32  \cdot  t^{2} = 4 + 64\cdot t - 16 \cdot  t^{2}.

6 0
3 years ago
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