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svetlana [45]
3 years ago
14

Planet A is an inner planet with no moon and hardly any atmosphere. Planet B is an inner planet with no moon but with a dense at

mosphere. Which pair of planets is being described?
SAT
2 answers:
lesya [120]3 years ago
5 0
Hey friend!
You stuck? 

There are two inner planets that have no moon. Mercury, and Venus. Only Mercury has hardly any atmosphere. Therefore, Planet A is Mercury. Venus which is composed primarily of carbon dioxide with a small amount of nitrogen is considered a planet with a dense atmosphere. Therefore, Planet B is Venus. 

Hope this helps! {Don't forget Brainliest!} :)
olganol [36]3 years ago
5 0

Answer: Planet A: Mercury.

Planet B: Venus.

Explanation: In our solar system there are only two planets without moons or natural satellites:

Mercury and Venus.

While Mercury has a very tenuous and variable atmosphere, Venus atmosphere is so dense that is almost impossible to see the surface of the planet, big clusters of clouds travel in the skies of Venus at really big speeds, reaching numbers like 350 km/h

So Mercury is the inner planet with no moon and almost no atmosphere, this means that Mercury is planet A, and Venus is the planet with no moon and a dense atmosphere, so Venus is planet B.

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TRANSCEND CLL 004 is a clinical trial that assesses a type of CAR T-cell therapy in patients suffering relapsed/refractory chronic lymphocytic leukemia/small lymphocytic lymphoma.

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In conclusion, TRANSCEND CLL 004 is a clinical trial that assesses a type of CAR T-cell therapy in patients suffering relapsed/refractory chronic lymphocytic leukemia/small lymphocytic lymphoma.

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8 0
2 years ago
What is 1 plus 1 in the correct for of The George Washington's formula?
sashaice [31]

Answer:

2 to the 1st power is the answer.

3 0
3 years ago
A projectile is projected from the origin with a velocity of 30
Sliva [168]

For this we will use kinematic equations.

We need the time of flight to compute the range.

With the given information, the most useful formula is:

Δy=v0yt−g2t2

Because our equation is taking only vertical motion into account,

v0y=30sin(45)

Because we are looking for the range of the object,

Δy=0

This leaves us with only one unknown:  t

0=30sin(45)t−g2t2

g2t2=30sin(45)t

g2t=30sin(45)

t=60gsin(45)

For horizontal distance, the formula we need is:

Δx=v0xt

Because we are taking into account the horizontal range,

v0x=30cos(45)

In substituting our derived values into the equation we can see that:

Δx=30cos(45)∗60gsin(45)

Range=30cos(45)∗60gsin(45)

6 0
3 years ago
Suppose a certain scale is not calibrated correctly, and as a result, the mass of any object is displayed as 0. 75 kilogram less
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The correlation between the actual masses of a set of objects and the respective masses of the same set of objects displayed by the scale is; 1

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We are told that a certain scale is not calibrated correctly,

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The scale of display = 1

Thus, the correlation between actual mass and displayed mass is 1 due to the fact that correlation is independent of change of origin and scale and correlation between any value and value minus 0.75 will be one.

Thus, the correlation between the actual masses of a set of objects and the respective masses of the same set of objects displayed by the scale is 1.

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4 0
3 years ago
a city of punjab has a 15 percent chance of wet weather on any given day. what is the probability that it will take a week for i
Marrrta [24]

Answer:

h

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h

4 0
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