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babymother [125]
3 years ago
12

I need help with this question. Thank you

Mathematics
2 answers:
bagirrra123 [75]3 years ago
7 0
Lwh/ 3 so its B. 120 ft ^3
Tatiana [17]3 years ago
4 0
B would be the answer
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I have a question,wheres the graph?

Step-by-step explanation:

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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
3 years ago
Someone please help
Gnoma [55]

Answer:

A, D, and E

Step-by-step explanation:

Triangles A, D, and E are the same size and shape, therefore congruent.

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3 years ago
Triangle UVW has a perimeter of 11.5 units. Solve for y.
Lina20 [59]

Answer:

The answer for y is 4 units

Step-by-step explanation:

The obvious question to ask in this situation is, “how many miles does Joseph travel on Mondays”? To compute, we each distance: 3 + 6 + 6 = 15.

Joseph travels 15 miles on Mondays.

Another way to work with this situation is to draw a shape that represents Joseph’s travel route and is labeled with the distance from one spot to another.

Notice that the shape made by Joseph’s route is that of a closed geometric figure with three sides (a triangle) (see figure 2). What we can ask about this shape is, “what is the perimeter of the triangle”?

Perimeter means “distance around a closed figure or shape” and to compute we add each length: 3 + 6 + 6 = 15

Our conclusion is the same as above: Joseph travels 15 miles on Mondays.

However, what we did was model the situation with a geometric shape and then apply a specific geometric concept (perimeter) to computer how far Joseph traveled.

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