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Setler [38]
3 years ago
11

I will mark Brainliest and I WILL REPORT LINKS!!!

Mathematics
2 answers:
jeka57 [31]3 years ago
4 0

Answer:

For

 y = (x^2 -4)/((x +2)(x^2 -49))

the numerator factors to (x -2)(x +2), so the factor of (x +2) will cancel with that in the denominator, leaving

 y = (x -2)/(x^2 -49)

There are points of discontinuity at the hole, x=-2, and at each of the vertical asymptotes, at x=-7, +7.

The horizontal asymptote is y=0.

Step-by-step explanation:

Fiesta28 [93]3 years ago
3 0

Answer:

Imma be honest here. I'm just answering so the other person can get brainliest

Step-by-step explanation:

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3x - 12 = -5 plus 12
3x-12 +12 = -5+12
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3x/3 = 7/3
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Twelve identical cylindrical pop cans are placed in a box. If sand fills the space between the pop cans and the cans and the sid
Masteriza [31]

Answer:

Step-by-step explanation:

I don't see any information on dimensions of the cans or box, so I'll assume the question wants a general solution.

Three steps will result in the volume of sand required to fill the remaining space in the box.

1.  Calculate the volume of a can using the equation for the volume of a cylinder:  Vol = πr²h

2.  Multiply the can volume by 12.  This will give the total volume of the 12 cans in the box.

3.  Calculate the volume of the box:  Volume = (Base)(Width)(Height)

[Make sure the units are the same as those used in calculating the can volume.  E.g., if the can is calculated with cm as the measue, the box dimensions must also be cm]

4.  Subtract the volume of the12 cans (from step 2) from the volume of the box.

The result from 4 will be the volume of sand required to fill the box containing 12 cans.  

I'm not clear why filling the box with sand is important, but perhaps it will absorb any soda released from cans broken when the heavy box is dropped by the person shocked that 12 cans of pop could be so heavy.

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8 0
2 years ago
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The width of a rectangle is 6 2/3 inches. The length of the is twice it’s width. What so the perimeter of the rectangle?
Nookie1986 [14]
\text {Width = }  6\dfrac{2}{3}  \text { inches}



The length is twice its width:
\text {Length = } 2 \times 6\dfrac{2}{3} \text { inches}

Change to improper fraction:
\text {Length = }2 \times \dfrac{20}{3} \text { inches}

Combine into single fraction:
\text {Length = } \dfrac{40}{3} \text { inches}


Find Perimeter :
\text {Perimeter = Length + Length + Width + Width}

\text {Perimeter = } \dfrac{40}{3} + \dfrac{40}{3}   + \dfrac{20}{3}   + \dfrac{20}{3} = \dfrac{120}{3}  = 40 \text { inches}


\bf \text {Answer: Perimeter =  40 inches}
6 0
4 years ago
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