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2 years ago

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.

Mathematics

1 answer:

Vlada [557]2 years ago

4 0

Answer:

It is symmetric about x- axis

The graph is not symmetric along the y - axis

It is not symmetric about the origin

Step-by-step explanation:

We can easily know the solution to your problem by graphing the equation

r(θ) = 2*cos (5*θ)

Please, see attached image

The graph is symmetric about the x axis if

r(θ) = r(-θ)

r(θ) = 2*cos (5*θ) = 2*cos (-5*θ) = r(-θ)

Since the cosine is an even function.

It is symmetric about x- axis

The graph is not symmetric along the y - axis, because

r(θ) ≠ r(π-θ)

It is symmetric about the origin if

r(θ) = r(π+θ)

But, since,

r(θ) = 2*cos (5*θ) ≠ 2*cos (5*(θ+π)) = 2*cos (5*θ+5*π)

It is not symmetric about the origin

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<u>Step-by-step explanation:</u>

The vertex form of a quadratic equation is: y = a(x - h)² + k

where (h, k) is the vertex ⇒ k is the maximum height

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maximum height (k) = 98 feet

distance traveled (x) = $\sqrt{\dfrac{-98}{-7.1}}+2.3$ = 6.02 seconds

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maximum height (k) = 112 feet

distance traveled (x) = $\sqrt{\dfrac{-112}{-7.5}}+2.2$ = 6.06 seconds

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maximum height (k) = 112 feet

distance traveled (x) = $\sqrt{\dfrac{-112}{-6.9}}+2.3$ = 6.33 seconds

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maximum height (k) = 95 feet

This has a lower height than the given equation.

D) H(t) = -7.5(t - 2.3)² + 95

maximum height (k) = 95 feet

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Let value intially be = P

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So after 1 year value is decreased by 0.2P

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so 20% of 0.8P = $\frac{20}{100} \times 0.8P = (0.2)(0.8P)$

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Similarly after 3 years, this value $P(0.8)^{2}$ will again be decreased by 20 %

so 20% of $P(0.8)^{2} \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}$

So after 3 years value is decreased by $(0.2)P(0.8)^{2}$

so value after 3 years will be = $P(0.8)^{2} - (0.2)P(0.8)^{2}$

taking $P(0.8)^{2}$ common out we get $P(0.8)^{2}(1-0.2)$

$P(0.8)^{2}(0.8)$

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so value after x years will be $P(0.8)^{x}$ ( whatever is the year, that is raised to power on 0.8)

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