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DanielleElmas [232]

4 years ago

13

What is 25 5/8 - 10 1/8

1 answer:

Delvig [45]4 years ago

4 0

Subtract 25 and 10 which gives you 15
subtract 5/8 and 1/8 since they have common denominators.
You get 15 1/2 when you simplify 15 4/8 in which 4/8 is the same thing as 1/2.

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The perimeter of a triangle is 199 inches. If one side of the triangle is six more than the shorter side, and the longer side is

Keith_Richards [23]

Shortest side (a) = 58
middle side (b) = 64
longest side (c) = 77

the 3 sides a + b + c = 199
b = a + 6
c = a + 19

substitute your new values for b & c into your original formula, so:
a + (a+6) + (a+19) = 199
3a + 25 = 199
3a = 174
a = 58

then substitute 58 into your b & c formulas to figure out the rest
b = a + 6 = 58 + 6 = 64
c = a + 19 = 58 + 19 = 77

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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