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iren [92.7K]
3 years ago
9

I need all questions please, will be giving brainliest and please make sure the answers are correct

Mathematics
1 answer:
Andrews [41]3 years ago
4 0
9 8 22
21 5 6
70 9 0
0 80 33
1 6 2
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F = (4x-3)au carre - (x+3)(3-9x)
Vinvika [58]

Answer:

<em>Translater</em>

Step-by-step explanation:

<em>F = (4x-3) squared - (x + 3) (3-9x)</em>

<em>- develop and reduce (4x-3) to the edge</em>

<em>- show that F = (5x) squared</em>

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A robot has a straight arm 20 inches long that can rotate about the origin of a coordinate system. If the robot’s hand is locate
scoundrel [369]

Answer:

The location of the robots hand after rotating 45° counterclockwise from (-20, 0) in inches is (-14.14, 14.14)

The location of the robots hand after rotating 45° clockwise from (-20, 0)) in inches is (-14.14, -14.14)

Step-by-step explanation:

The initial location of the robots hand = (-20, 0)

The angle through which the robot rotates his hand = 45°

∴ We have the length of the robots hand = 20 inches

We note that (-20, 0) is in the 2nd Quadrant

The location of the robots hand after rotating 45° counterclockwise = 135°

Therefore, the location of the robots hand after the rotation = (20×cos(135°), 20×sin(135°)) = (-10·√2, 10·√2) = (-14.14, 14.14)

The location of the robots hand after rotating 45° clockwise from (-20, 0) = 225°

Therefore, the location of the robots hand after the rotation = (20×cos(225°), 20×sin(225°)) = (-10·√2, -10·√2) = (-14.14, -14.14)

8 0
3 years ago
Solve the following equation with the initial conditions. x¨ + 4 ˙x + 53x = 15 , x(0) = 8, x˙ = −19
Katen [24]

x''+4x'+53x=15

has characteristic equation

r^2+4r+53=0

with roots at r=-2\pm7i. Then the characteristic solution is

x_c=C_1e^{(-2+7i)t}+C_2e^{(-2-7i)t}=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)

For the particular solution, consider the ansatz x_p=a_0, whose first and second derivatives vanish. Substitute x_p and its derivatives into the equation:

53a_0=15\implies a_0=\dfrac{15}{53}

Then the general solution to the equation is

x=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)+\dfrac{15}{53}

With x(0)=8, we have

8=C_1+\dfrac{15}{53}\implies C_1=\dfrac{409}{53}

and with x'(0)=-19,

-19=-2C_1+7C_2\implies C_2=-\dfrac{27}{53}

Then the particular solution to the equation is

\boxed{x(t)=\dfrac1{53}e^{-2t}(409cos(7t)-27\sin(7t)+15)}

8 0
3 years ago
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