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2 years ago

What's the highest common factor of 5y and y^2?

Mathematics

2 answers:

dangina [55]2 years ago

4 0

Hello.

The highest common factor of

$\mathrm{5y} \:and\:y^{2}$

$y$

y is the greatest common factor (G.C.F.) also called the Highest Common Factor (HCF)

Therefore, y is the highest common factor of 5y and y².

I hope it helps.

Have a nice day.

$\boxed{imperturbability}$

lys-0071 [83]2 years ago

3 0

<h3>Answer: y</h3>

This is because y is a factor of 5y and it's also a factor of y^2 = y*y

It's the largest common factor between the two given expressions.

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4 years ago

The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.

VLD [36.1K]

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The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

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To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

$\int\limits^2_{-1} {(f(x) - g(x))} \, dx$

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx$

We know that:

$\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}$

$\int\limits^{}_{} {1} \, dx = x$

$\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}$

Then our integral is:

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2} + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3} + (-1) - \frac{(-1)^2}{2} )$

The right side is equal to:

$(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6$

The bounded area is 5 + 5/6 square units.

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3 years ago