Help me find the measure of the arc!!!!!!

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

7 0

90 + 146= 90+146=236 which you subtract from 360 so therefore you’re answer would be 124 degrees

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10 points please answer correctly (due soon)

luda_lava [24]

The exponential equation of the model is A(t) = 2583 * 0.88^t and the multiplier means that the number of new cases in a week is 88% of the previous week

<h3>The function that models the data</h3>

The given parameters are:

New, A(t) = 2000

Rate, r = 12%

The function is represented as:

A(t) = A * (1 - r)^t

So, we have:

2000 = A * (1 - 12%)^t

This gives

2000 = A * (0.88)^t

2 weeks ago implies that;

t = 2

So, we have:

2000 = A * 0.88^2

Evaluate

2000 = A * 0.7744

Divide by 0.7744

A = 2583

Substitute A = 2583 in A(t) = A * 0.88^t

A(t) = 2583 * 0.88^t

Hence, the exponential equation of the model is A(t) = 2583 * 0.88^t

<h3>The interpretation of the multiplier</h3>

In this case, the multiplier is 88% or 0.88

This means that the number of new cases in a week is 88% of the previous week

Read more about exponential equation at

brainly.com/question/2456547

#SPJ1

8 0

2 years ago

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$