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nekit [7.7K]
3 years ago
12

Write the next 3 numbers in this pattern. Then describe the rule for this pattern in an nth expression.

Mathematics
1 answer:
hram777 [196]3 years ago
8 0

Answer:

An=10^n + 1 is the expression for the nth term.

Next three terms:

100001

1000001

10000001

Step-by-step explanation:

The number ot 0's imcrease between the two 1's starting at no zeros, 1 zero, two zeros, and so on...

So the next terms are 100001, 1000001, 10000001 and that's without finding the formula.

Now let's find the formula:

A1=11=10+1

A2=101=100+1

A3=1001=1000+1

Notice we have increasing powers of 10 and then just adding 1.

A1=10^1 +1

A2=10^2 +1

A3=10^3 +1

....

An=10^n +1

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Please Help!!!!!!! I need it!
sergiy2304 [10]

Answer:

Rotation - Twists the shape

Reflection - Flips the shape

Translation - Slides a shape

Dilation - Enlarge or reduces a shape

Congruent - Same shape same size

Rigid Motion - type of transformation that  creates congruent figures

Hope this helped. I apologize if any of them are wrong..

5 0
3 years ago
Read 2 more answers
I would like all the help I can’t get pleaseee
expeople1 [14]

Answer:

Step-by-step explanation:

a=4    b=12

2a+3b=

2*4+3*12=

8+36=

44

a²-b+10=

4²-12+10=

4*4-2=

16-2=

14

(ab)²-(a+b)²=

(4*12)²-(4+12)²=

48²-16²=

(48+16)(48-16)=

64*32=

2048

7 0
2 years ago
What is the area of the figure below?
irinina [24]
1/2 x b x h = 36 , like if u find that please :) .
4 0
3 years ago
To dip or not to diP
Aleks04 [339]

Answer:

Dip shunnnnnnnnnnnnnn

Step-by-step explanation:

4 0
3 years ago
Prove :
Galina-37 [17]
Figure 1 shows the triangle ABC with the angle bisectors AD,           BE and CF of its three angles A, B and C respectively. The points D, E and F are the intersection points of the angle bisectors and the opposite triangle sides. 
Since the straight lines AD and BE are the angle bisectors to the angles A and B respectively, they can not be parallel, otherwise the sides AB and BC would be in one straight linewhat is not the case. Therefore, the straight lines AD and BE intersect in some point P. 
From the lesson An angle bisector properties (Theorem 1) we know that the points of the angle bisector AD are equidistant from the sides AB and AC of the angle BAC. 

Figure 1. To the Theorem              

Figure 2. To the proof of the TheoremIn particular, the point P is equidistant from the sides AB and AC of the angle BAC. This means that the perpendiculars GP and HP (Figure 2) drawn from the point P to the sides AB and AC are of equal length: GP = HP. 
By the same reason, the points of the angle bisector BE are equidistant from the sides AB and BC of the angle ABC. In particular, the point P is equidistant from the sides AB and BC of the angle ABC. This means that the perpendiculars GP and IP (Figure 2) drawn from the point P to the sides AB and BC are of equal length: GP = IP. 
Two equalities above imply that the perpendiculars HP and IP are of equal length too: HP = IP. In other words, the point P is equidistant from the sides AC and BC of the angle ACB. In turn, it implies that the intersection point P lies at the angle bisector CF of the angle ACB in accordance to the Theorem 2 of the lesson An angle bisector properties. In other words, the angle bisector CF of the angle ACB passes through the point P. 
Thus, we have proved that all three angle bisectors DG, EH and FI pass through the point P and have this point as their common intersection point.Since the point P is equidistant from the triangle sides AB, BC and AC, it is the center of the inscribed circle of the triangle ABC (Figure 2).So, all the statements of the Theorem are proved. 
The proved property provides the way of constructing an inscribed circle for a given triangle.To find the center of such a circle, it is enough to construct the angle bisectors of any two triangle angles and identify their intersection point. This intersection point is the center of the inscribed circle of the triangle. To get the radius of the inscribed circle you should to construct the perpendicular from the found center of the inscribed circle to any triangle side. 
5 0
3 years ago
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