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eduard
3 years ago
12

When astronauts Neil Armstrong and Buzz Aldrin first walked on

Mathematics
2 answers:
MAVERICK [17]3 years ago
7 0

Answer:

16.5 pounds

Step-by-step explanation:

16.5% as a decimal is 0.165

0.165 x 100 = 16.5

So in conclusion you will weigh about 16.5 pounds on the moon.

marissa [1.9K]3 years ago
3 0
About 16 if we are counting the fact that the moons gravity is about 16.5 of the earth you will weight 16.5 pounds and will be able to float!
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Sin2x = 2 sinx cosx
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3 years ago
If x is an integer, then is -x positive?
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If x is an integer, then for values of x ≤ 0 would -x be positive.

General Formulas and Concepts:

<u>Math</u>

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Step-by-step explanation:

We know that integers comprise of the number line from -∞ to ∞. We can have numbers like -3, -2, -1, 0, 1, 2 ,3.

If we say that x is an integer, and that -x must be positive, then that means the integer x must be negative, because a negative times a negative is a positive.

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3 years ago
What value of x makes the equation true?
soldier1979 [14.2K]
This one is easy.
<span>–8.6 = –4.1 + x
</span>
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X=-4.5
The answer is B
3 0
4 years ago
Read 2 more answers
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harkovskaia [24]

Answer:

4,055

Step-by-step explanation:

7 0
3 years ago
(a) Show that a differentiable function f decreases most rapidly at x in the direction opposite the gradient vector, that is, in
Sophie [7]

Answer:

Step-by-step explanation:

\text{Show that a differentiable function f decreases most rapidly at x in the }

\text{direction opposite the gradient vector, that is, in the direction of} -\bigtriangledown f(x)\text{. Let}\  \theta \ \text{be the angle between} \bigtriangledown f(x) \  \text{and unit vector u. Then } D_u f = \mathbf{|\bigtriangledown f| \  cos  \ \theta }}

\text{Since the minimum value of} \ \  \mathbf{cos   \ \theta} \  \ is \mathbf{-1} \  \text{occuring \ for \ 0} \le \ \theta \ < 2x,  \\ \\ when  \ \theta = \mathbf{\pi} , \text{the mnimum value of} \  D_uf  \ is} -|\bigtriangledown f|,  \text{occuring when the direction of u is } \\ \\  \ \mathbf{the \ opposite \  of} \  \text{the direction of }  \ \bigtriangledown f (assuming \ \bigtriangledown f\ is \  not \ zero)

b) \text{From part A:}

If \ f(x,y) = x^4y -x^2y^2 \ \  decreases \ fastest \ at \ the \point \ (2,-5)\\ \\ F(x,y) = x^4y -x^2y^3 \\ \\ f_x = \dfrac{df}{dx}= \dfrac{d}{dx}(x^4y-x^2y^3)  \\ \\ f_x = \dfrac{df}{dx}= y4x^3 -2y^3x  \\ \\ For(2,-5) \\ \\ f_x = (-5)4(2)^3 -2(-5)^3(2) \\ \\ \mathbf{ f_x = 340}

However; f_y = \dfrac{df}{dy} = \dfrac{d}{dy}(x^4y - x^2y^3) \\ \\ f_y = x^4 -3x^2y^2 \\ \\  Now, for (2, -5)\\ \\f_y = (2)^4 -3(2)^2(-5)^2 \\ \\ f_y = -284

So; \bigtriangledown = < 340,-284> \text{this is the direction of fastest decrease}

6 0
3 years ago
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