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Olegator [25]
3 years ago
14

Brainliest answer

Mathematics
1 answer:
Flura [38]3 years ago
5 0

A geometric sequence is a sequence of numbers that follows a pattern where the next term is found by multiplying by a constant called the common ratio, q:

a_n=a_{n-1}\cdot q=(a_{n-2}\cdot q)\cdot q=a_{n-2}\cdot q^2=...,\\ a_n=a_1\cdot q^{n-1}.

Given the functions f(n)=11 and g(n)=\left(\dfrac{3}{4}\right)^{n-1}, you can consider a_n=11\cdot \left(\dfrac{3}{4}\right)^{n-1} as n-th term of the geometric sequence with first term a_1=11 and common ratio q=\dfrac{3}{4}.

For n=9,

a_9=11\cdot \left(\dfrac{3}{4}\right)^{9-1} =11\cdot \left(\dfrac{3}{4}\right)^8=1.101.

Answer: the correct choice is B.

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1)

Given the triangle RST with Coordinates  R(2,1), S(2, -2) and T(-1 , -2).

A dilation is a transformation which produces an image that is the same shape as original one, but is different size.  

Since, the scale factor \frac{5}{3} is greater than 1, the image is enlargement or a stretch.  

Now, draw the dilation image of the triangle RST with center (2,-2) and scale factor \frac{5}{3}

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ST=3, so S{}'T{}'=5  

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Since, RT=3\sqrt{2} [By using hypotenuse of right angle triangle] and R{}'T{}'=5\sqrt{2}.


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Disagree with the given statement.

Side Angle Side postulate (SAS) states that:

If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle then these two triangles are congruent.

Given: B is the midpoint of \overline{AC} i.e \overline{AB}\cong \overline{BC}

In the triangle ABD and triangle CBD, we have

\overline{AB}\cong \overline{BC}   (SIDE)            [Given]

\overline{BD}\cong \overline{BD}   (SIDE)            [Reflexive post]

Since, there is no included angle in these triangles.

∴ \Delta ABD is not congruent to \Delta CBD .

Therefore, these triangles does not follow the SAS congruence postulates.

(b)

SSS(SIDE-SIDE-SIDE) states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

Since it is also given that  \overline{AD}\cong \overline{CD}.

therefore, in the triangle ABD and triangle CBD, we have

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\overline{AD}\cong \overline{CD}   (SIDE)           [Given]

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therefore by, SSS postulates \Delta ABD\cong \Delta CBD.

3)

Given that:  \angle1=\angle 3 are vertical angles, as they are formed by intersecting lines.

Therefore

, by the definition of linear pairs

\angle 1 and \angle 2 and \angle 3  and \angle 2 are linear pair.

By linear pair theorem, \angle 1 and \angle 2   are supplementary, \angle 2 and \angle 3  are supplementary.

m\angle1+m\angle 2=180^{\circ}

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Equate the above expressions:

m\angle 1+m\angle 2=m\angle 2+m\angle 3

Subtract the angle 2 from both sides in the above expressions

∴m\angle 1=m\angle 3

By Congruent Supplement theorem: If two angles are supplements of the same angle, then the two angles are congruent.


therefore, \angle 1\cong \angle 3.















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