I’ll bet you nobody’s not gonna answer this question

Mathematics

2 answers:

Arisa [49]3 years ago

6 0

Answer:

Lol this is my first time seeing someone saying I bet no one will answer this. Ok coming back to the topic answers are :-

Initial value = -18

Initial value is the output value or the y variable. Looking in the table the y variable is -18 since x is 0.

Rate of change = 0.5

You have to use the formula of rate of change to find the answer. You can choose any pairs from the table and will give you the same answer. I pick the first pair and the last pair to find my rate of change.

(0, -18) and (36, 0)

x1, y1 x2, y2

Rate of change = y2 - y 1 / x2 - x1

Rate of change = 0 - (-18) / 36 - 0

Rate of change = 18 / 36

Rate of change = 0.5

Hope this helps, thank you :) !!

stich3 [128]3 years ago

3 0

Hi im here to answer the question now i won the bet

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Fred orders pictures from a photographer. Each picture costs $5.50. A one-time shipping fee of $4.25 is added to the cost of the

Alexus [3.1K]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

Read 2 more answers

Am i correct? calculus

Advocard [28]

Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

$\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
\\\\\\
\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\
$

$\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}$