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2 years ago

Is (x+1)^2 equivalent to x^2 + 1? explain

Mathematics

1 answer:

Mademuasel [1]2 years ago

4 0

Answer:

No.

Step-by-step explanation:

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

This is different from x^2+1.

Thus, they are not equivalent.

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Solve the system of linear equations by substitution.

Goryan [66]

Answer:

hey! This was the only part of mathematics I ever enjoyed! So consider this both of our lucky days: Answer is : (2,7)

Step-by-step explanation:

basically plug in Y to the first equation as sox + 4(-x + 9) = 30

x - 4x + 36 = 30- 3x = - 6

x = 2

then plug that in to Y!

Y = -(2) + 9

so Y = 7

and our answer is : (2,7)

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Match the following terms by evaluating the expression for x=-6

Scorpion4ik [409]

The sum of 2 opposites =0
X+6 would now equal 0
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3 years ago

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Nat2105 [25]

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3 years ago

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Elena L [17]

Answer:

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3 years ago

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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

Angelina_Jolie [31]

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

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

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4 years ago