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3 years ago

Compute the missing data in the table for the following exponential function . Leave your answer in decimal form.

Mathematics

1 answer:

Snezhnost [94]3 years ago

4 0

Answer: Option C: 0.03125

Step-by-step explanation:

we have an exponential equation, so we have:

f(x) = A*r^x

we have that:

f(1) = A*r^1 = A*r = 0.5

f(2) = A*r^2 = 0.25

f(3) = A*r^3 = 0.125

f(4) = A*r^4 = 0.0625

f(5) = A*r^5 = ?

we want to find the value f(5)

first, let's take the quotient:

f(2)/f(1) = (A*r^2)/(A*r) = r = 0.25/0.5 = 0.5.

then we have:

f(x) = A*0.5^x

and f(1) = A*0.5 = 0.5

then A = 0.5/0.5 = 1.

now we know that our function is f(x) = 0.5^x.

then, if we want to find f(5) we have:

f(5) = 0.5^5 = 0.03125

Then the correct option is option c.

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Identify the standard form of the equation by completing the square.

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Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

Given equation:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$

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nevsk [136]

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4 wands = 6 rands -> 6*4=24 so 4*4=16 -> 16 wands = 24 rands (1 wand=1.5 rand)

24 rands = 8 fands -> 24/8 = 3 -> 3 rands = 1 fand

3 rands means 1.5*3=4.5 -> 4.5 wands = 1 fand -> 4.5*5=22.5 wands

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