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tatiyna
4 years ago
15

What is the commom difference in the following arithmetic sequence??

Mathematics
1 answer:
aalyn [17]4 years ago
5 0
Each increases by a certain amount called the common difference

4.4-2.8=1.6
6-4.4=1.6
7.7-6=1.7
hmm, some error
maybe you meant 7.6

common difference is 1.6

answer is C
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If A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7}, what is A U B?
Phoenix [80]

Answer:

Answer is A U B = {3,5,7}

3 0
3 years ago
givings direccións to find different ranger staton.unluckily all the ranger are busy rescuing others campers and can't come to y
DENIUS [597]
To find the resultant vector you need to add the horizontal and vertical components of each individual vector.
Cosine for horizontal, Sine for vertical.
v = \ \textless \ |v|cos \theta, |v| sin \theta\ \textgreater \
Where |v| is magnitude or length of vector.
Lets look at first ranger station. First vector has magnitude 4.5 with angle of 90.
Next has magnitude of 8.1 with angle of 125.
F_1 = \ \textless \  4.5 cos(90)+8.1 cos(125) , 4.5 sin(90) + 8.1  sin(125) \  \textgreater \
F_1 = \ \textless \ -4.646, 11.135\ \textgreater \
Find magnitude:
|F_1| =  \sqrt{((-4.646)^2+(11.135)^2} = 12.065
Do the same with 2nd ranger station:
F_2 = \ \textless \  7.5 cos(20) + 5.3 cos(100), 7.5 sin(20) + 5.3 sin(100)\ \textgreater \
F_2 = \ \textless \ 6.127, 7.784\ \textgreater \
|F_2| = \sqrt{(6.127)^2 + (7.784)^2}  = 9.9

Therefore the 2nd ranger station is closest to starting point.

6 0
4 years ago
Urgent need assistance
pashok25 [27]
I'm here because you asked for assistance.
3 0
3 years ago
Geometry help please
ExtremeBDS [4]
It looks about 90 degrees
8 0
4 years ago
A regular polygon is defined to be a(n) _____ polygon with congruent sides and congruent angles
Solnce55 [7]

Answer:

Step-by-step explanation:

A regular polygon is defined to be a convex polygon with congruent sides and congruent angles.

Convex means that all interior angles are less than 180 degrees.  However, if all interior angles are equal, the polygon has to be convex.

5 0
4 years ago
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