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Lunna [17]
2 years ago
6

Which is a right triangle formed using a diagonal through the interior of the cube?

Mathematics
2 answers:
grigory [225]2 years ago
3 0

Answer:

triangle AEH is da good one

Step-by-step explanation:

Free_Kalibri [48]2 years ago
3 0

Answer:

The correct option is triangle GDC

Step-by-step explanation:

The dimensions give for the cube are such that the top surface has vertices GBCF while the bottom surface has vertices HADE.

A right angle can be formed in quite a number of ways since the cube has right angles on all six surfaces. However the question states that the diagonal that forms the right angle runs "through the interior."

Therefore option 1 is not correct since the diagonal formed in triangle BDH passes through two surfaces. Triangle DCB is also formed with its diagonal passing only along one of the surfaces. Triangle GHE is also formed with its diagonal running through one of the surfaces.

However, triangle GDC is formed with its diagonal passing through the interior as shown by the "zigzag" line from point G to point D. And then you have another line running from vertex D to vertex C.

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\$13,107.96

Step-by-step explanation:

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4 years ago
Part 1: What mistake did AJ make in the graph?
grigory [225]

Answer:

Part 1) AJ drawn the parabola opening upwards, instead of drawing it opening downwards

Part 2) see the explanation

Step-by-step explanation:

Part 1) What mistake did AJ make in the graph?

we have

f(x)=-(x+2)^2-1

This is the equation of a vertical parabola written in vertex form

The parabola open downward (because the leading coefficient is negative)

The vertex represent a maximum

The vertex is the point (-2,-1)

therefore

AJ drawn the parabola opening upwards, instead of drawing it opening downwards

Part 2) Evaluate any two x-values (between -5 and 5) into AJ's function. Show your work. How does your work prove that AJ made a mistake in the graph?

take the values x=-4 and x=4

For x=-4

substitute the value of x in the quadratic equation

f(x)=-(-4+2)^2-1\\f(x)=-5

For x=4

substitute the value of x in the quadratic equation

f(x)=-(4+2)^2-1\\f(x)=-37

According to AJ's graph for the value of x=-4 the function should be positive, however it is negative and for the value of x=4 the function should be positive and the function is negative

therefore

AJ made a mistake in the graph

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Step-by-step explanation:

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