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avanturin [10]
3 years ago
15

A space shuttle orbits Earth one time in 90 minutes. How many times does orbit earth in six hours?

Mathematics
2 answers:
pshichka [43]3 years ago
5 0
90 minutes =1.5 hours=one orbit
Divide 6 by 1.5 because this will help you see how many orbits can be made in 6 hours (since 1.5 hours=1 orbit)
6/1.5=4
Therefore the space shuttle orbits around the Earth 4 times in 6 hours.
Hope this helps!
Otrada [13]3 years ago
3 0
6 divided by 1.5 = 4

The shuttle orbits earth 4 times in 6 hours.
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jasenka [17]

The linear equation is y = 4x - 4

And the graph of the linear equation can be seen in the image below.

<h3>How to graph the last line?</h3>

It seems that you already are good at graphing, so I will try to explain how to find the equation more in detail.

Remember that a general linear equation is written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

In this case, we know that the y-intercept is -4, then b = -4, replacing that we get:

y = a*x - 4

Now we also can see that this line passes through the point (2, 4), this means that if we evaluate in x = 2, the outcome should be y = 4, replacing that we get:

4 = a*2  - 4

4 + 4 = a*2

8 = a*2

8/2 = a = 4

Then the slope is 4, and the linear equation is:

y = 4x - 4

The graph is below.

If you want to learn more about linear functions:

brainly.com/question/4025726

#SPJ1

8 0
1 year ago
Complex to Trigonometric:<br>Z=√6+√2+2i​
Alekssandra [29.7K]

Answer:

z=6+2i maybe

Step-by-step explanation:

8 0
3 years ago
Use a truth table to show that P Qand (~PV Q) A (~QV P) are equivalen
kati45 [8]

Answer:  The given logical equivalence is proved below.

Step-by-step explanation:  We are given to use truth tables to show the following logical equivalence :

P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P)

We know that

two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.

The truth table is as follows :

P     Q      ∼P     ∼Q     P⇔ Q    ∼P ∨ Q     ∼Q ∨ P        (∼P ∨ Q)∧(∼Q ∨ P)

T     T         F        F             T            T                   T                       T

T     F         F        T             F             F                   T                       F

F     T         T        F             F            T                   F                       F

F     F         T        T             T            T                   T                       T

Since the corresponding truth vales for P ⇔ Q and (∼P ∨ Q)∧(∼Q ∨ P) are same, so the given propositions are logically equivalent.

Thus, P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P).

8 0
3 years ago
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Gwar [14]
Assuming this is some kind of modular question, maybe mod 17.

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Evgen [1.6K]
That would be 6/b <====
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