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3 years ago

(2 weeks, 4 days, 17 hours, 30 minutes) + 3

SAT

2 answers:

Komok [63]3 years ago

7 0

Answer:

33 minutes i think

Explanation:

might be becuase im wrong lol

ohaa [14]3 years ago

6 0

Is 68 I hope it helps

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A spool is on a horizontal surface (with friction) and pulled to the right with a thread attached to the center of the spool, so

IceJOKER [234]

The relationship between the magnitude of the force of tension and the force of friction is; Ff = (r/R)Ft

<h3>What is Rotational Acceleration?</h3>

We are given;

mass of spool = M

Radius = R

moment of inertia = I.

Now, the spool is not moving and as such;

r < R which means there is net torque = 0

Thus;

Ff × R = Ft × r

>> Ff = (r/R)Ft

Finally, the angular acceleration is expressed as;

α = a/R

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What values of c and d would make the following expression represent a real number? i(2 3i)(c di) c = 2, d = 3 c = –2, d = 3 c =

Olegator [25]

Complete question:

<em>What values of c and d would make the following expression represent a real number? i(2 + 3i)(c + di) </em>

<em>choices: A)c = 2, d = 3 B)c = –2, d = 3 C)c = 3, d = –2 D)c = –3, d = –2</em>

<em />

<em>Option D</em><em> is correct. </em> The value of c and d for the expression to be real are -3 and -2 respectively

Given the expression

$i(2 + 3i)(c + di)$

Expand the expression

$=i(2 + 3i)(c + di)\\=(2i+3i^2)(c+di)\\=(2i+3(-1))(c+di)\\=(2i-3)(c+di)\\=2ic+2di^2-3c-3di\\=2ic-2d-3c-3di$

Collect the like terms

$=2ic-3di-2d-3c$

For the resulting function to be real, then the imaginary part must be zero i.e.

$2ic-3di=0\\2ci = 3di\\2c = 3d\\c=\frac{3d}{2}\\$

If d = -2, then;

$c=\frac{3(-2)}{2}\\c=\frac{-6}{2}\\c= -3$

Hence the value of c and d for the expression to be real are -3 and -2 respectively.

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