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3 years ago

Rewrite the function by completing the square. f(x)= x^{2} -9 x +14f(x)=x 2 −9x+14

Mathematics

1 answer:

Bess [88]3 years ago

5 0

Answer:

$f(x)=(x-4.5)^{2}-6.25$ or $f(x)=(1/4)(2x-9)^{2}-6.25$

Step-by-step explanation:

we have

$f(x)=x^{2}-9x+14$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-14=x^{2}-9x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-14+4.5^{2}=x^{2}-9x+4.5^{2}$

$f(x)+6.25=x^{2}-9x+4.5^{2}$

Rewrite as perfect squares

$f(x)+6.25=(x-4.5)^{2}$

$f(x)=(x-4.5)^{2}-6.25$

$f(x)=(1/4)(2x-9)^{2}-6.25$

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Find the equation that models the path of a satellite if its path is a hyperbola, a=55,000 km and c=81,000 km. Assume the center

Yuki888 [10]

Given:

Hyperbola

a=55,000 km and c=81,000 km

hyperbola is the origin and the transverse axis is horizontal

Required:

Equation of the path of a satellite

Solution:

Formula for hyperbola, (x-h)2/a2 – (x-k)2/b2 = 1

At origin, (h, k) = (0, 0)

(x-0)2/(55000)2 – (x-0)2/(81000)2 = 1

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2 years ago

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3[ (5)/(-9) (6)/(-2)]-2[(0)/(-2)(2)/(-4)]

Anika [276]

5 is the correct answer I believe

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2 years ago

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At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

Answer-

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

Solution-

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

6 0

2 years ago

Y is between points p and b. pt= 15 and tb = 10. what is pb

Kruka [31]

PT = 15, and TB = 10

Take the 15 from PT, and add the 10 from TB. Which equals 25.

PB = 25

So, 25 is your answer.

Hope this helps! ☺

4 0

3 years ago

Y-2x + 1 How to do it

patriot [66]

Answer:

It is in simplest form so you leave it like that

Step-by-step explanation:

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2 years ago

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