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3 years ago

Rewrite the complex number in polar form

Mathematics

1 answer:

salantis [7]3 years ago

8 0

r = sqrt ( (-2)^2 + (2sqrt3)^2)

= sqrt (4 + 12) = 4

tan theta = 2 sqrt 3 / -2 = -sqrt3

theta = 2pi/3

The correct option is C

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The height in meters of a toy rocket at t seconds can be found by the

OlgaM077 [116]

Answer: 81.2

Step-by-step explanation:

h(t) = -5t² + 40t + 1.2

h(4) = -5(4)² + 40(4) + 1.2 We plug 4 seconds as t, and solve h(4)

h(4) = -80 + 160 + 1.2

h(4) = 80 + 1.2

h(4) = 81.2

4 0

3 years ago

The radius of a right circular cone is increasing at a rate of 1.4 in/s while its height is decreasing at a rate of 2.8 in/s. At

zepelin [54]

Answer:

34191.7πin³/sec

Step-by-step explanation:

Volume of circular come = 1/3pi x r² x h

When we differentiate this formula we have

dv/dt = 1/3π[r²dh/dt + 2rhdr/dt]

We have the following information

r = 138

H = 143 inch

dr/dt = 1.4

dh/dt = -2.8

When we plug into the formula

1/3π[130² x -2.8 + 2x138x143x1.4]

= 34191.7πin³/sec

7 0

3 years ago

An alloy of silver and gold is to be recast with an addition of silver. This new alloy consists of 80 ounces and is to have only

Liula [17]

Answer:

What percentage of gold is in the new alloy after the recasting? 4

Step-by-step explanation:

(80/5)*(1/5)/80=0.04

5 0

3 years ago

Help with 1 and 2 please

Anna35 [415]

Answer:

love your nails, can u bring it a bit closer tho?

Step-by-step explanation:

4 0

2 years ago

An ice cube is melting, and the lengths of its sides are decreasing at a rate of 0.8 millimeters per minute At what rate is the

julia-pushkina [17]

Answer:

The rate of decrease is: $43.2mm^3/min$

Step-by-step explanation:

Given

$l = 18mm$

$\frac{dl}{dt} = -0.8mm/min$ ---- We used minus because the rate is decreasing

Required

Rate of decrease when: $l = 18mm$

The volume of the cube is:

$V = l^3$

Differentiate

$\frac{dV}{dl} = 3l^2$

Make dV the subject

$dV = 3l^2 \cdot dl$

Divide both sides by dt

$\frac{dV}{dt} = 3l^2 \cdot \frac{dl}{dt}$

Given that: $l = 18mm$ and $\frac{dl}{dt} = -0.8mm/min$

$\frac{dV}{dt} = 3 * (18mm)^2 * (-0.8mm/min)$

$\frac{dV}{dt} = 3 * 18 *-0.8mm^3/min$

$\frac{dV}{dt} = -43.2mm^3/min$

<em>Hence, the rate of decrease is: 43.2mm^3/min</em>

8 0

3 years ago