Leah took her friends for a ride in her hot air balloon. The function fff models the height of the hot air balloon above the ground (in meters) as a function of time (in minutes) after takeoff.
Plot the point on the graph of fff that corresponds to when the hot air balloon landed.

Answers

Answer 1

Answer: point (60,0)

Step-by-step explanation: the x-intercept (60,0) shows that at 60 minuetes the balloon is 0 meter above the ground

Answer 2

The coordinates of the function of height and time are (210, 0) and (60, 10). The graph is attached below.

What is function?

The function is a relationship between a set of potential outputs and a set of possible inputs, where each input has a single relationship with each output. This means that if an object x is present in the set of inputs (also known as the domain), then a function f will map that object to exactly one object f(x) in the set of potential outputs (called the co-domain).

Given:

r(t) = 210 - 15t

Calculate the height after 10 minutes as shown below,

The height = 210 - 15 × 10

The height = 210 - 150

The height = 60

At time 0 the height will be,

The height = 210 - 15 × 0

The height = 210 - 0

The height = 210 meters,

Thus, the coordinates will be (210, 0) and (60, 10)

To know more about function:

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Leah Took Her Friends For A Ride In Her Hot Air Balloon. The Function Fff Models The Height Of The Hot

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[tex] \: \: \: \: \: \: [/tex]

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16 is 80% of what number?
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Answers

Answer:

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Step-by-step explanation:

1. To solve this problem, instead of doing 80/100 times 16, we can do 100/80 times 16 (inverse of percentage, basically).

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Therefore, the answer is A. 20.

find the point (x,y) on the curve where dy/dx = 0

Answers

a. By the chain rule,

[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} = \dfrac{dy}{dt}\cdot\dfrac{1}{\frac{dx}{dt}}[/tex]

Given that [tex]y=e^t+e^{-t}[/tex] and [tex]x=e^{-t}[/tex], we have the derivatives

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and so

[tex]\dfrac{dy}{dx} = \dfrac{e^t - e^{-t}}{-e^{-t}} = 1-e^{2t}[/tex]

Set this equal to zero and solve for t :

[tex]1-e^{2t} = 0[/tex]

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[tex]\ln(1) = \ln\left(e^{2t}\right)[/tex]

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This value of t corresponds to x = e⁰ = 1 and y = e⁰ - 1/e⁰ = 1 - 1 = 0. So the only point on the curve where the derivative dy/dx is zero is (1, 0).

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[tex]\dfrac{d^2}{dx^2} = \dfrac d{dx} \dfrac{dy}{dx} = \dfrac{df}{dx} = \dfrac{df}{dt}\cdot\dfrac{dt}{dx} = \dfrac{df}{dt}\cdot\dfrac{1}{\frac{dx}{dt}}[/tex]

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[tex]\left(\dfrac{d^2y}{dx^2}\right)^2 + \dfrac{dy}{dx} - 1 = 0[/tex]

[tex]\left(2e^{3t}\right)^2 + (1-e^{2t}) - 1 = 0[/tex]

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[tex]\ln\left(e^{4t}\right) = \ln\left( \dfrac14\right)[/tex]

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[tex]4t = -\ln(4)[/tex]

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Answer:

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Step-by-step explanation:

hope this helps!

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Answer:

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Step-by-step explanation:

correct me if im wrong(┬┬﹏┬┬)

Complete the statement based on the following information.

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Answer:

  (c)  ΔTRS

Step-by-step explanation:

Perhaps the easiest way to find corresponding vertices is to look at the angle markings:

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Then, in order of corresponding angles, we have ...

  ΔJKI ≅ ΔTRS

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Step-by-step explanation:

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Answer:479001600

Step-by-step explanation:

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A way to do it by hand is

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Answer:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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A(-5, -4) → A' is a glide reflection where the translation is (x, y) → (x + 6, y), and the line of reflection is y = 3. What are the coordinates of A'?

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Answers

Answer:

(1, -4)

Step-by-step explanation:

(-5,-4) -> (x,y) so x = -5 and y = -4. Apply the rule (x+6,y) to that point by substituting in your x and y values.

(-5+6,-4) = (1, -4)

(Express these ratios in their simplest form)
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BRAINLIEST**

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Answer:

3:4 itself.

3:4 itself.

Step-by-step explanation:

3:4 itself.

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Answers

Answer:

[tex]-\frac{x(15x^4-2x^3+3)}{3}[/tex]

Step-by-step explanation:

[tex]\frac{30x^8-4x^7+6x^4}{-6x^3}[/tex]

The first thing I would do is to factor the numerator. All coefficients there are divisible by 2, and all variables are divisible by x⁴. I'll also move the negative sign to the left.

[tex]-\frac{2x^4(15x^4-2x^3+3)}{6x^3}[/tex]

Next, you can simplify 2/6 to 1/3:

[tex]-\frac{x^4(15x^4-2x^3+3)}{3x^3}[/tex]

Finally, the last thing you can do is simplify x⁴/x³ to x/0:

[tex]-\frac{x(15x^4-2x^3+3)}{3}[/tex]

That's as simplified as it gets.

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Answers

Answer:

Neither.

Step-by-step explanation:

Manipulate the formulas into slope-intercept form which is [tex]y=mx+b[/tex].

In order for them to be parallel, m needs to be the same. Perpendicular, m for one needs to be the negative reciprocal of the other [tex]m=-\frac{1}{m}[/tex].

Let's see what we get.

(1)

[tex]3x+4y=2\\4y=-3x+2\\y=-\frac{3}{4}x+1/2[/tex]

(2)

[tex]4x+3y=2\\3y=-4x+2\\y=-\frac{4}{3}x+2/3[/tex]

They are reciprocals, but they are both negative. So, they are not the same line, perpendicular, nor parallel.

Also, you can graph these equations on desmos and see easily that they have none of these relationships.

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-2 1/2 - (-1 3/4)

When subtracting a negative vile the equation becomes addition:

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Answer: -3/4

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