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miss Akunina [59]
3 years ago
11

2 less than four times a number is equal to twice a number increase by 8 write an equation and solve to find the value of x

Mathematics
1 answer:
Tom [10]3 years ago
5 0
4x - 2 = 2x + 8
2x -2 = 8
2x = 10
x = 5

4(5)-2 = 2(5)+8
20-2 = 10+8
18 = 18
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Solve the differential equation 3ydx+(xy+5x)dy=0
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The ODE is separable:

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Integrating both sides gives

\displaystyle\int\left(1+\frac5y\right)\,\mathrm dy=-3\int\frac{\mathrm dx}x\implies y+5\ln|y|=-3\ln|x|+C

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374.123 m^2

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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]

Answer:

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).

We want to find the area bounded by:

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

g(x) = y = x

x = -1

x = 2

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
Use the zeros of the following quadratic to find the x-value of the vertex.
Ivan
B I got this by graphing it into my graphing calculator. Hope this helps
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