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Vlada [557]
2 years ago
14

What is the relationship between APB and minor arc AB?

Mathematics
1 answer:
Dmitry [639]2 years ago
4 0

Answer:

Step-by-step explanation:

If APB is measured in radians, then the arc length AB = APB(r) where r is the radius of the circle.

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Eman is planning to sell wind chimes at a craft fair. The cost of her tools for building the wind chimes is 130. The cost of mat
antoniya [11.8K]

We are given : The cost of tools for building wind chimes = $130.

Cost of material for each wind chime = $10.

We need to find the total cost of making 100 wind chimes.

Please note : $130 is the fix cost and $10 is the variable cost of each wind chime.

<em>In order to find the cost of 100 wind chimes, we need to multiply 100 by cost of making one wind chime and add the fix cost for tools for building wind chimes.</em>

Therefore, total cost of making 100 wind chimes = 10×100 + 130

= 1000+130

=1130.

<h3>Therefore, the total cost to make 100 wind chimes is $1130.</h3>

7 0
2 years ago
What is the anser for 194*33333
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(the answer is) = 6466602

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One bag of rice that weighs 5.25 pounds will have enough rice if a recipe calls for 88 ounces of rice. (1 pound = 16 ounces)
krek1111 [17]
False, there isn't enough rice in 5.25 pounds.
4 0
3 years ago
Jordan has $55. She earns $67 by doing chores. How much money does jordan have now?
Nuetrik [128]
If Jordan has $55 and she earns $67 by doing chores, you can calculate how much money does Jordan have now using the following step:

$55 + $67 = $122

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6 0
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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} }

4 0
2 years ago
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