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2 years ago

15

40 is the sum of 15 and Craig's savings. Use the variable c to represent Craig's savings.

2 answers:

andrew-mc [135]2 years ago

8 0

40=15+c

Hope it helps

levacccp [35]2 years ago

8 0

The answer to this will be 40=15+c and this is because it’s sag 40 is the sum. With that we know that 40 is how they have in total and the word sum indicates that it’s addition. After that it says of 15 and Craig’s saving. We know that it’s addition so we can make it 15+c and then your equation is complete with 40=15+c which also means the Craig had 25 dollars

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3 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]

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.

3 0

2 years ago