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marishachu [46]
3 years ago
6

I need help please !

Mathematics
1 answer:
earnstyle [38]3 years ago
5 0
No it’s not because a polynomial has a 6 not a 2
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Describe the pattern, write the next term, and write a rule for the nth term of the sequence,
vagabundo [1.1K]

Step-by-step explanation:

Next term is - 51

Use formula Tn=a+(n-1)d

a=the first term of the sequence which is - 15

d=the difference between the terms which is - 9

Substitute

Tn=-15+(n-1)(-9)

Tn=-15+9-9n

Tn=-9n-6

From this you will be able to generate all the terms in the pattern

If they say determine the 50th term just substitute 50 by n

Tn=-9(50)-6

Tn=-450-6

Tn=-456

7 0
3 years ago
Given: PRST square
yuradex [85]
There is no difficulty in this problem until you construct the figures. How can we do it is shown in the attached picture. After drawing PRST, from the point P, we can draw PMKD and later we can complete PMCT as a result. From this picture, we can see that the side of PMCT is also a. Then, the area of this square is A=a^{2}

6 0
3 years ago
Which statements about the local maximums and minimums for the given function are true? Choose three options.
Nostrana [21]

Answer:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.  

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum  is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

8 0
3 years ago
Read 2 more answers
Which is the pair of congruent right angles?
Alex73 [517]

The little squares at corner-B and corner-E were drawn there
to show that those are right angles.

8 0
3 years ago
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The points where a graph crosses the x- and y-axis are called the _____.
Marina CMI [18]
I think it's called a coordinate plane


Sorry if I'm wrong
6 0
3 years ago
Read 2 more answers
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