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3 years ago

What is the correct answer?

Physics

1 answer:

pentagon [3]3 years ago

3 0

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

- x^3-4x^2 <-- (x-4)(x^2)

_________________

6x^2-16x-32

- 6x^2-24x <-- (x-4)(6x)

_________________

8x-32

- 8x-32 <- (x-4)(8)

___________________________

0 | x^2+6x+8

This means the answer is B) x^2+6x+8

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Thanks for answering

ElenaW [278]

Answer:

II) Kitchen waste: Meal leftovers, Banna peelings

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III) A

IV) Home: Bottles of shampoo, leftover food, syringe

office Gloves

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Laboratory: empty cartridge

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2 years ago

In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t

Tamiku [17]

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

$V_a$ = Velocity of airplane in still air

$V_w$ = Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

$V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)$

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

$V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)$

Subtracting the two equations we get

$V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h$

Applying the value of velocity of wind to the first equation

$V_a-40=460\\\Rightarrow V_a =500\ km/h$

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

5 0

3 years ago

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

Rudik [331]

Answer:

1) P₁ = -2 D, 2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

$\frac{1}{f_1} = \frac{1}{q}$

q = f₁

2) for an object located at p = 25 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}$

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}$

$\frac{1}{f_2}$ = 0.06

f₂ = 16.67 cm

the expression for the power of the lenses is

P = $\frac{1}{f}$

where the focal length is in meters

1) P₁ = 1/0.50

P₁ = -2 D

2) P₂ = 1 /0.16667

P₂ = 6 D

4 0

2 years ago

A car has a mass of 1000 kg and accelerates at 2 meters per second per second. what is the magnitude of the net force exerted on

9966 [12]

F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N

4 0

3 years ago

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Use the SI prefixes in Table 3 of this chapter to convert these hypothetical units of measure into appropriate quantities: a. 10

8090 [49]

A. 10 rations = 1 deca-ration.

b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.

c. 10⁻⁵ phones = 1 micro-phones.

d. 10⁻⁹ goats = 1 nano-goats.

e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.

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2 years ago