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Mariulka [41]
3 years ago
15

Which of the following stanements shows the inverse property of addition?

Mathematics
1 answer:
Nikitich [7]3 years ago
4 0

Answer:

0\cdot \:y+\left(-0\cdot \:y\right)=0  the  statement shows the inverse property of addition as the sum of a number and its inverse is zero.

Step-by-step explanation:

As we know that Inverse Property of Addition states if you add any number to its opposite, the result will be zero.

For example, 13 + (-13) = 0 shows that -13 is the additive inverse of 13.

In other words, the sum of a number and its inverse is zero.

Now checking from the available options,

0y+\left(-0y\right)

\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a

=0\cdot \:y-0\cdot \:y

\mathrm{Add\:similar\:elements:}\:0\cdot \:y-0\cdot \:y=0

=0

Thus,

0\cdot \:y+\left(-0\cdot \:y\right)=0

Therefore, 0\cdot \:y+\left(-0\cdot \:y\right)=0  the  statement shows the inverse property of addition as the sum of a number and its inverse is zero.

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Find the volume of the wedge with vertices at points (0,0,0), (1,0,0), (0,1,0), (0,0,1) by integrating the area of cross-section
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Answer:

V = 1/6 cubic units

Step-by-step explanation:

Applying the concept of integrals for volume calculation:

V = \int\limits^b_a {S(x)} \, dx          (1)

V = volume of the solid bounded by x = a and x = b

S(x) = cross section area of the solid, perpendicular to the x axis

From the figure we have that S is the area of a triangle that has base Z and height Y

Area of the triangle = S(x)=\frac{y(x)*z(x)}{2}          (2)

Calculation of y(x) and z(x)

We apply the equation of the point-slope line (plane xy):

Slope = m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}          (3)

Equation of the line = y - y_{1} =m(x-x_{1} )          (4)

Replacing the points (1,0) and (0,1) in (3):

m=\frac{1-0}{0-1} =-1

Replacing the point (1,0) and m = -1 in (4):

y-0=(-1)(x-1)

y(x) = -x + 1 (Line A-B)          (5)

We apply the equation of the point-slope line (plane xz):

Slope = m = \frac{z_{2} - z_{1} }{x_{2} - x_{1}}          (6)

Equation of the line = z - z_{1} =m(x-x_{1} )          (7)

Replacing the points (1,0) and (0,1) in (6):

m=\frac{1-0}{0-1} =-1

Replacing the point (1,0) and m = -1 in (7):

z-0=(-1)(x-1)

z(x) = -x + 1 (Line A-C)        (8)

Replacing (5) and (8) in (2)

S(x) = \frac{(-x + 1) * (-x + 1)}{2} =\frac{(-x + 1)^{2} }{2}          (9)

Replacing (9) in (1) and knowing that a = 0 and b = 1:

V = \int\limits^1_0 {\frac{(-x + 1)^{2} }{2}} \, dx = \int\limits^1_0 {\frac{x^{2}-2x+1 }{2}} \, dx

V =\frac{1}{2} (\frac{x^{3} }{3} -2\frac{x^{2} }{2} +x)  evaluated from x=0 to x=1

V= \frac{1}{2} (\frac{1}{3} -1 +1) = \frac{1}{6}

3 0
3 years ago
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