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3 years ago

I need to know how these triangles are congruent

Mathematics

1 answer:

ZanzabumX [31]3 years ago

3 0

Answer:

They both are right triangles and they both have tick marks which is showing that the sides are congruent

Step-by-step explanation:

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Write the series using summation notation.<br><br> 4 + 8 + 12 + 16 + 20+ . . . + 80

zhannawk [14.2K]

Answer:

$\sum ^{20} _{n \to \11} 4n$

Step-by-step explanation:

In order to write the series using the summation notation, first we need to find the nth term of the sequence formed. The sequence generated by the series is an arithmetic sequence as shown;

4, 8, 12, 16, 20...80

The nth term of an arithmetic sequence is expressed as Tn = a +(n-1)d

a is the first term = 4

d is the common difference = 21-8 = 8-4 = 4

n is the number of terms

On substituting, Tn = 4+(n-1)4

Tn = 4+4n-4

Tn = 4n

The nth term of the series is 4n.

Since the last term is 80, L = 4n

80 = 4n

n = 80/4

n = 20

This shows that the total number of terms in the sequence is 20

According to the series given 4 + 8 + 12 + 16 + 20+ . . . + 80 , we are to take the sum of the first 20terms of the sequence. Using summation notation;

4 + 8 + 12 + 16 + 20+ . . . + 80 = $\sum ^{20} _{n \to \11} 4n$

4 0

3 years ago

given the recursive formula for a geometric sequence find the common ratio the 8th term and the explicit formula.did I set these

lesya [120]

Answer:

Step-by-step explanation:

1)Since we know that recursive formula of the geometric sequence is

$a_{n}=a_{n-1}*r$

so comparing it with the given recursive formula $a_{n}=a_{n-1}*-4$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=-2*(-4)^{7} =32768.$

Explicit Formula = $-2*(-4)^{n-1}$

2) Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

8th term= $a_{1}*(r)^{n-1}=-4*(-2)^{7} =512.$

Explicit Formula = $-4*(-2)^{n-1}$

3)Comparing the given recursive formula $a_{n}=a_{n-1}*3$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =3

8th term= $a_{1}*(r)^{n-1}=-1*(3)^{7} =-2187.$

Explicit Formula = $-1*(3)^{n-1}$

4)Comparing the given recursive formula $a_{n}=a_{n-1}*-4$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=3*(-4)^{7} =-49152.$

Explicit Formula = $3*(-4)^{n-1}$

5)Comparing the given recursive formula $a_{n}=a_{n-1}*-4$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=-4*(-4)^{7} =65536.$

Explicit Formula = $-4*(-4)^{n-1}$

6)Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

8th term= $a_{1}*(r)^{n-1}=3*(-2)^{7} =-384.$

Explicit Formula = $3*(-2)^{n-1}$

7)Comparing the given recursive formula $a_{n}=a_{n-1}*-5$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

8th term= $a_{1}*(r)^{n-1}=4*(-5)^{7} =-312500.$

Explicit Formula = $4*(-5)^{n-1}$

8)Comparing the given recursive formula $a_{n}=a_{n-1}*-5$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

8th term= $a_{1}*(r)^{n-1}=2*(-5)^{7} =-156250.$

Explicit Formula = $2*(-5)^{n-1}$

6 0

3 years ago

Sistema de ecuaciones.5x+2y=-152x-2y=-6

Zolol [24]

Tnemos el sisema de ecuaciones:

$\begin{gathered} 5x+2y=-15 \\ 2x-2y=-6 \end{gathered}$

Podemos resolverlo por eliminación sumando ambas ecuaciones y eliminando y. Asi podemos resolver para x:

$\begin{gathered} (5x+2y)+(2x-2y)=(-15)+(-6) \\ 7x+0y=-21 \\ x=-\frac{21}{7} \\ x=-3 \end{gathered}$

Ahora podemos resolver para y con cualquiera de las dos ecuaciones:

$\begin{gathered} 2x-2y=-6 \\ 2\cdot(-3)-2y=-6 \\ -6-2y=-6 \\ -2y=-6+6 \\ -2y=0 \\ y=0 \end{gathered}$

Respuesta: x=-3, y=0

7 0

1 year ago

laura purchased a prepaid phone card for $25 . long distance calls cost 8 cents a minute using this card. laura used her card on

ira [324]

Amount spent on the last call = $25 - $22.04 = $2.96 = 296 cents
Number of minutes used for last call = 296 / 8 = 37 minutes.

6 0

3 years ago

Using the until circle, what is the exact value of tan 2pi/3

Klio2033 [76]

Tan 2 pi/3= (sin2pi/3)/(cos2pi/3)= (sqrt 3)/2 dividend by (-1/2)= - sqrt 3

7 0

3 years ago

Read 2 more answers

I need to know how these triangles are congruent​

I need to know how these triangles are congruent