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3 years ago

Find the diameter, approximate circumference and area of the circle below. (Use 3 for π)

Mathematics

2 answers:

vovikov84 [41]3 years ago

5 0

Answer:

Circumference = 42

Area = 147 to the power of 2

Step-by-step explanation:

7x2=14 diameter

3x14=42 circumference

Area = 7x7x3 =147 to the power of 2

seraphim [82]3 years ago

4 0

Diameter=14
Circumference =43.98
Area=153.94

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Answer:

divide 500 by 125.

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2 years ago

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Arturiano [62]

Answer:

Step-by-step explanation:

Option C is the correct answer.

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3 years ago

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

At what point on the parabola y = 3x^2 + 2x is the tangent line parallel to the line<br> y = 10x −2?

oksian1 [2.3K]

Answer:

( $\frac{4}{3}$ , 8 )

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 10x - 2 ← is in slope- intercept form

with slope m = 10

Parallel lines have equal slopes

then the tangent to the parabola with a slope of 10 is required.

the slope of the tangent at any point on the parabola is $\frac{dy}{dx}$

differentiate each term using the power rule

$\frac{d}{dx}$ (a $x^{n}$ ) = na $x^{n-1}$ , then

$\frac{dy}{dx}$ = 6x + 2

equating this to 10 gives

6x + 2 = 10 ( subtract 2 from both sides )

6x = 8 ( divide both sides by 6 )

x = $\frac{8}{6}$ = $\frac{4}{3}$

substitute this value into the equation of the parabola for corresponding y- coordinate.

y = 3( $\frac{4}{3}$ )² + 2

= (3 × $\frac{16}{9}$ ) + 2

= $\frac{16}{3}$ + $\frac{8}{3}$

= $\frac{24}{3}$

= 8

the point on the parabola with tangent parallel to y = 10x - 2 is ( $\frac{4}{3}$ , 8 )

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2 years ago

What is the relationship between 1.0 and 0.1

dexar [7]

1/10×1.0=0.1

answer: C

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3 years ago