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3 years ago

Solve for the value of z. (z+2)° (4z-7)°

Mathematics

1 answer:

telo118 [61]3 years ago

3 0

Answer:

Z=0.8

Step-by-step explanation:

Step-by-step explanation:

Combine the opposite terms

2z+2+4z=7

Add

2z and 4z

6z+2=7

Move all terms not containing z to the right side of the equation.

Subtract 2 from both sides of the equation.

6z=7−2

6z=5

Divide each term:

6z/6 = 5/6

Decimal Form:

z=0.8

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Please explain step by step on how you got the answer.

Jobisdone [24]

The answer is 21

First you po perenthies which would be 8-6=2 the equation now is 5+4×(2)2

Then you to exponents which would be 2×2=4 the equation now is 5+4×4

Now you multiply 4×4=16 the equation is now 5+16

Now you add 5 and 16 to get 21

7 0

3 years ago

The equation r(t) = sin(4t)i + cos(4t)j, 0t≥0 describes the motion of a particle moving along the unit circle. Answer the follo

lorasvet [3.4K]

Answer:

a) Particle has a constant speed of 4, b) Velocity and acceleration vector are orthogonal to each other, c) Clockwise, d) False, the particle begin at the point (0,1).

Step-by-step explanation:

a) Let is find first the velocity vector by differentiation:

$\vec v = \frac{dr_{x}}{dt} i + \frac {dr_{y}}{dt} j$

$\vec v = 4\cdot \cos 4t\, i - 4 \cdot \sin 4t \,j$

$\vec v = 4 \cdot (\cos 4t \, i - \sin 4t\,j)$

Where the resultant vector is the product of a unit vector and magnitude of the velocity vector (speed). Velocity vector has a constant speed only if magnitude of unit vector is constant in time. That is:

$\|\vec u \| = 1$

Then,

$\| \vec u \| = \sqrt{\cos^{2} 4t + \sin^{2}4t }$

$\| \vec u \| = \sqrt{1}$

$\|\vec u \| = 1$

Hence, the particle has a constant speed of 4.

b) The acceleration vector is obtained by deriving the velocity vector.

$\vec a = \frac{dv_{x}}{dt} i + \frac {dv_{y}}{dt} j$

$\vec a = 16\cdot (-\sin 4t \,i -\cos 4t \,j)$

Velocity and acceleration are orthogonal to each other only if $\vec v \bullet \vec a = 0$ . Then,

$\vec v \bullet \vec a = 64 \cdot (\cos 4t)\cdot (-\sin 4t) + 64 \cdot (-\sin 4t) \cdot (-\cos 4t)$

$\vec v \bullet \vec a = -64\cdot \sin 4t\cdot \cos 4t + 64 \cdot \sin 4t \cdot \cos 4t$

$\vec v \bullet \vec a = 0$

Which demonstrates the orthogonality between velocity and acceleration vectors.

c) The particle is rotating clockwise as right-hand rule is applied to model vectors in 2 and 3 dimensions, which are associated with positive angles for position vector. That is: $t \geq 0$