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3 years ago

What is R when N = 31.5ft and slope = 4/12

Mathematics

1 answer:

Alexxandr [17]3 years ago

3 0

MAN i suck at this stufff.... I'll try to find someone who will help tho

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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

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

3 0

3 years ago

Find the logarithmic function y = logbx that passes through the points

katovenus [111]

$b\in(0,\ 1)\ \cup\ (1,\ \infty)\\x > 0\\y\in\mathbb{R}\\--------------------------\\\\y=\log_bx\\\\For\ (1,\ 0)\to x=1,\ y=0.\ Substitute:\\\\\log_b1=0\to b^0=1\to b\in(0,\ 1)\ \cup\ (1,\ \infty)\\\\For\ (116,\ 2)\to x=116,\ y=2.\ Substitute:\\\\\log_b116=2\to b^2=116\to b=\sqrt{116}\\\to b=\sqrt{4\cdot29}\to b=\sqrt4\cdot\sqrt{29}\to b=2\sqrt{29}\\\\For\ (4,\ -1)\to x=4,\ y=-1.\ Substitute:\\\\\log_b4=-1\to b^{-1}=4\to b=\dfrac{1}{4}$