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vova2212 [387]
3 years ago
15

How to solve -7-3x+2=-8x-8

Mathematics
2 answers:
murzikaleks [220]3 years ago
5 0
I think it's this i'm not sure x= - 5/3 

help me with my question pls
viktelen [127]3 years ago
5 0
First, you have to put the numbers by its kind so you have to change it to
-7+2+8=-8x+3x
then just add and subtract
3=-5x
x= -3/5
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Yes 11/13 cannot be simplified anymore
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lara [203]
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What are the values of x and y
Murljashka [212]

E 10 12

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4 0
2 years ago
the second term of a geometric sequence is 18 and the fourth term is 8 find the common ratio . And find the sum of the first 6 t
777dan777 [17]

Answer:

  • ratio: 2/3
  • sum: 73 8/9

Step-by-step explanation:

The general term of a geometric sequence is ...

  an = a1·r^(n-1)

You have the 2nd and 4th terms, so ...

  a2 = a1·r^(2-1) = a1·r

  a4 = a1·r^(4-1) = a1·r^3

We can find r from the ratio ...

  a4/a2 = (a1·r^3)/(a1·r) = r^2 = 8/18 = 4/9

Then r is ...

  r = √(4/9) = 2/3 . . . . the common ratio

The first term is ...

  a2 = 18 = a1·(2/3)

  a1 = (3/2)·18 = 27

__

The sum of the first 6 terms is ...

  Sn = a1·(r^n -1)/(r -1)

  S6 = 27·((2/3)^6 -1)/(2/3 -1)

  S6 = 27·(64/729-1)/(2/3-1) = (27)(665)/243 = 73 8/9

The sum of the first 6 terms is 73 8/9.

_____

<em>Check on the sum</em>

The first 6 terms are ...

  27, 18, 12, 8, 5 1/3, 3 5/9

Their sum is 73 8/9, as above.

8 0
3 years ago
Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(4, 0), Q(0, −4), and R(−8, −4). Triangle P'Q'R' has
Whitepunk [10]

Answer:

Please find attached the required plot accomplished with an online tool

Part A:

1/4

Part B:

P''(-1, 0),  Q''(0, -1), and R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent

Step-by-step explanation:

Part A:

Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)

Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)

The dimensions of the sides of the triangle are given by the relation;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where;

(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment

For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);

By substitution into the length equation, we get;

The length of segment PQ = 4·√2

The length of segment PR = 4·√10

The length of segment RQ = 8  

The length of segment P'Q' = √2

The length of segment P'R' = √10

The length of segment R'Q' = 2

Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4

Part B:

Reflection of (x, y) across the y-axis gives;

(x, y) image after reflection across the y-axis = (-x, y)

The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;

P'(1, 0) image after reflection across the y-axis = P''(-1, 0)

Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)

R'(-2, -1) image after reflection across the y-axis = R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.

6 0
3 years ago
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