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nika2105 [10]
3 years ago
6

A force F1 of magnitude 6.00 units acts at the origin in a direction 30.0 above the positive x axis. A second force F2 of magnit

ude 5.00 units acts at the origin in the direction of the positive y axis. Find graphically the magnitude and direction of the resultant force F1 + F2
Mathematics
1 answer:
kumpel [21]3 years ago
5 0
F1 = 6*cos(30) i + 6*sin(30) j = 5.20 i + 3 j 
<span>F2 = 0 i + 5 j </span>

<span>R = F1 + F2 </span>
<span>Add the 'i's together (x direction) and the 'j's together (y direction) </span>
<span>R = (5.20 + 0) i + (3 + 5) j = 5.2 i + 8 j </span>

<span>Magnitude: </span>
<span>||R|| = </span>√<span>(5.2^2 + 8 ^2) = </span>√<span>91 = 9.54 </span>

<span>Direction: </span>
Θ<span> = atan ( Rj / Ri ) = atan( 8 / 5.2 ) = 57.0</span>⁰
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Step-by-step explanation:

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Vladimir79 [104]

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8 4/15

Step-by-step explanation:

Rewriting our equation with parts separated

= 5 + 2/3 + 2 + 3/5

Solving the whole number parts

5 + 2 = 7

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2/3 + 3/5 = ?

Find the LCD of 2/3 and 3/5 and rewrite to solve with the equivalent fractions.

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3 years ago
1. ) Consider the function f(x)=5−7x2,−5≤x≤1
Marat540 [252]
1.) The interval of the value of x is from -5 to 1, inclusive. Remember that what is asked is the absolute value, thus the sign does not matter even if you have to subtract x from 5. Thus, the maximum value would be obtained if the x is smaller, which is 1. The minimum value is obtained when x=-5.

Absolute maximum value: x = - 5
f(-5) = ║5 - 7(-5)^2║ = ║-170║=170


Absolute minimum value: x = 1
f(1) = ║5 - 7(1)^2║ = ║-2║= 2

2.) The Mean Value Theorem (MVT) applies to functions that are continuous and differentiable on the closed and open interval of a to b, respectively. Since the function is a quadratic function, MVT can be applied. Then, this means that there is a value of c which is between a and b. This could be determined using this formula according to MVT:

f'(c)= \frac{f(b)-f(a)}{b-a}

The differentiated form would be f'(x) = -2x. Then,

-2c =  \frac{(4- 0^{2} )-(4- (-1)^{2}) }{0--1}=1

c=- \frac{1}{2}

Thus, x = -1, x = -1/2, and x=0 all lie in the function 4-x^2.
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