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wlad13 [49]
2 years ago
13

A student estimates the weight of astronauts on the moon by multiplying their weight by the decimal 0.16666....what fraction can

be used for the same estimation?
Mathematics
2 answers:
Y_Kistochka [10]2 years ago
8 0

Answer:

\frac{1}{6}      

Step-by-step explanation:

0.16666 can be written as \frac{16666}{1000000}

write the numerator and denominator as the product of their prime numbers

and divide the common numbers. Thus, it can be further reduced to \frac{1}{6}

The gravity of moon is about \frac{1}{6} times the gravity of the Earth due to which the weight of the astronauts on the moon can be found by multiplying by the decimal 0.16666.. or \frac{1}{6}.

Elden [556K]2 years ago
3 0

Let x = 1.66666... ....(1)

Multiply both sides of equation (1) by 10.

10x = 1.666666.... .... (2)

Multiply both sides of equation (1) by 100.

100x = 16.66666....... (3)

Subtract equation (2) from equation (3)

100x - 10x = 16.66666...- 1.66666...

90x = 15

x= \frac{15}{90}  = \frac{1}{6}

Hence, the required fraction form is \frac{1}{6} .... Answer.


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Leona [35]

Answer:

a. 5 b. y = -\frac{3}{4}x + \frac{1}{2} c. 148.5 d. 1/7

Step-by-step explanation:

Here is the complete question

Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons, and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit. Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point. Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f() is a real number Let f be an increasing function with f(0) = 2. The derivative of f is given by f'(x) = sin(πx) + x² +3. (a) Find f" (-2) (b) Write an equation for the line tangent to the graph of y = 1/f(x) at x = 0. (c) Let I be the function defined by g(x) = f (√(3x² + 4). Find g(2). (d) Let h be the inverse function of f. Find h' (2). Please respond on separate paper, following directions from your teacher.

Solution

a. f"(2)

f"(x) = df'(x)/dx = d(sin(πx) + x² +3)/dx = cos(πx) + 2x

f"(2) = cos(π × 2) + 2 × 2

f"(2) = cos(2π) + 4

f"(2) = 1 + 4

f"(2) = 5

b. Equation for the line tangent to the graph of y = 1/f(x) at x = 0

We first find f(x) by integrating f'(x)

f(x) = ∫f'(x)dx = ∫(sin(πx) + x² +3)dx = -cos(πx)/π + x³/3 +3x + C

f(0) = 2 so,

2 = -cos(π × 0)/π + 0³/3 +3 × 0 + C

2 = -cos(0)/π + 0 + 0 + C

2 = -1/π + C

C = 2 + 1/π

f(x) = -cos(πx)/π + x³/3 +3x + 2 + 1/π

f(x) = [1-cos(πx)]/π + x³/3 +3x + 2

y = 1/f(x) = 1/([1-cos(πx)]/π + x³/3 +3x + 2)

The tangent to y is thus dy/dx

dy/dx = d1/([1-cos(πx)]/π + x³/3 +3x + 2)/dx

dy/dx = -([1-cos(πx)]/π + x³/3 +3x + 2)⁻²(sin(πx) + x² +3)

at x = 0,

dy/dx = -([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)⁻²(sin(π × 0) + 0² +3)

dy/dx = -([1-cos(0)]/π + 0 + 0 + 2)⁻²(sin(0) + 0 +3)

dy/dx = -([1 - 1]/π + 0 + 0 + 2)⁻²(0 + 0 +3)

dy/dx = -(0/π + 2)⁻²(3)

dy/dx = -(0 + 2)⁻²(3)

dy/dx = -(2)⁻²(3)

dy/dx = -3/4

At x = 0,

y = 1/([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)

y = 1/([1-cos(0)]/π + 0 + 0 + 2)

y = 1/([1 - 1]/π + 2)

y = 1/(0/π + 2)

y = 1/(0 + 2)

y = 1/2

So, the equation of the tangent at (0, 1/2) is

\frac{y - \frac{1}{2} }{x - 0} = -\frac{3}{4}  \\y - \frac{1}{2} = -\frac{3}{4}x\\y = -\frac{3}{4}x + \frac{1}{2}

c. If g(x) = f (√(3x² + 4). Find g'(2)

g(x) = f (√(3x² + 4) = [1-cos(π√(3x² + 4)]/π + √(3x² + 4)³/3 +3√(3x² + 4) + 2

g'(x) = [3xsinπ√(3x² + 4) + 18x(3x² + 4) + 9x]/√(3x² + 4)

g'(2) = [3(2)sinπ√(3(2)² + 4) + 18(2)(3(2)² + 4) + 9(2)]/√(3(2)² + 4)

g'(2) = [6sinπ√(12 + 4) + 36(12 + 4) + 18]/√12 + 4)

g'(2) = [6sinπ√(16) + 36(16) + 18]/√16)

g'(2) = [6sin4π + 576 + 18]/4)

g'(2) = [6 × 0 + 576 + 18]/4)

g'(2) = [0 + 576 + 18]/4)

g'(2) = 594/4

g'(2) = 148.5

d. If h be the inverse function of f. Find h' (2)

If h(x) = f⁻¹(x)

then h'(x) = 1/f'(x)

h'(x) = 1/(sin(πx) + x² +3)

h'(2) = 1/(sin(π2) + 2² +3)

h'(2) = 1/(sin(2π) + 4 +3)

h'(2) = 1/(0 + 4 +3)

h'(2) = 1/7

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