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arsen [322]
3 years ago
9

Write the number 78 as a product of its prime factors

Mathematics
2 answers:
Gre4nikov [31]3 years ago
6 0
2/3 = .66666....
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...

Otrada [13]3 years ago
5 0
78|2\\ 39|3 \\ 13|13 \\ 1 \\ \\ 78=2*3*13
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What is lim x→-3 sqrt x^2-8
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If the  -8 is under the square root, then...

\displaystyle L = \lim_{x\to -3} \sqrt{x^2-8}\\\\L = \sqrt{(-3)^2-8}\\\\L = \sqrt{9-8}\\\\L = \sqrt{1}\\\\L = 1\\\\

OR

If the -8 is not under the square root, then...

\displaystyle L = \lim_{x\to -3} \sqrt{x^2}-8\\\\L = \sqrt{(-3)^2}-8\\\\L = \sqrt{9}-8\\\\L = 3-8\\\\L = -5

Either way, we replace x with -3 and simplify.

For more information, refer to the direct substitution rule for limits.

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2 years ago
The following list gives the number of pets for each of 9 students.
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3 years ago
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6 0
3 years ago
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The sum of 4 times a number and 15 is 47. Find the number.
Bezzdna [24]

Answer:

The number is 8.

Step-by-step explanation:

Write an equation then solve by isolating the variable.

let x be the number

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A gravel path is to be built around an 18‘ x 15‘ rectangle garden if the width remains constant how why can the path be if there
Ilia_Sergeevich [38]

Answer:

P = 2*(18 + 13) = 62ft

Now to figure out the width of the path we have to do some geometry. There are 3 basic shapes we will use for the path, they are the 13 ft long rectangle, the 18 ft long rectangle, and the squares formed at the corners. These squares' areas are equal to the width squared, whereas the rectangles are only equal to the length times the width. Our total area then becomes:

A = 2*(13*w) + 2*(18*w) + 4*(w*w)

A = 4w^2 + 62w

And since we know the available area is 516 ft^2, putting this in for A and solving the quadratic equation we can get the width:

516 = 4w^2 + 62w

0 = 4w^2 + 62w - 516

Using the quadratic formula: (w = -b2+/-SQRT(b^2 - 4ac)/2a, where a = 4, b = 62, and c = -516)

w = 6, -21.5

Since we obviously cannot have a negative width, 6 must be our answer. Therefore the path can be 6 feet wide!

We can check this answer by doing the following. Please note that by having a 6ft wide path you would actually be adding 12 ft to both dimensions, since you add 6 feet of length to both sides. Your total area would then become:

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Subtracting the area of the garden gives you the are of the gravel required:

750ft^2 - (18*13)ft^2 = 516ft^2

Therefore a width of 6 feet is correct.

Step-by-step explanation:

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3 years ago
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