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kobusy [5.1K]
3 years ago
5

Join our among us Code: KFTJFF

Mathematics
2 answers:
Pavlova-9 [17]3 years ago
8 0
Ok bet I tryyyyy!!!!!
MA_775_DIABLO [31]3 years ago
6 0

Answer:

ok

Step-by-step explanation:

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JUST ANSWER PLEASE!!! QUICK
jeka57 [31]

Answer:

<u>Options 1 and 3</u>

Step-by-step explanation:

We should know that, the system of linear equations can be treated as matrices, i.e: we can modify or make any operations provide that we must apply the same operation for all terms of each equation.

Given:  the solution for the following system is (2,9)

Px + Qy = R  ⇒(1)

Tx + Uy = V  ⇒(2)

We will check which system of equation has the same solution.

<u>System A)</u>  Px + Qy = R

                  (P+T)x + (Q+U)y = R+V  ⇒(3)

So, By summing (1) and (2) we will get the equation (3)

So, system A has the same solution (2,9)

<u>System B)</u> Px + Qy = R

                 (P+2T)x + (Q+2U)y = R-2V  ⇒(4)

By multiplying equation (2) by 2 and add with equation (1), we will get:

 (P+2T)x + (Q+2U)y = R+2V

Which is not the same as equation (4)

So, system B has not the same solution (2,9)

<u>System C)</u> (T-P)x + (U-Q)y = V-R  ⇒(5)

                  Tx + Uy = V  

By multiplying equation (1) by -1 and add with equation (2), we will get the equation (5)

So, system C has the same solution of (2,9)

<u>System D)</u> (T-P)x + (Q+U)y = V-R  ⇒(6)

                  Tx + Uy = V  

We cannot get equation (6) by the same operation over equation (1)

Note the coefficient of x and y⇒ (T-P) and (Q+U)

They must be (T+P) and (Q+U) <u>OR </u>(T-P) and (Q-U)

So, system D has not the same solution of (2,9)

<u>System E)</u> (5T-P)x + (5U-Q)y = V-5R ⇒ (6)

                  Tx + Uy = V  

By subtract equation (1) from 5 times equation (2), we will get:

(5T-P)x + (5U-Q)y = 5V-R

Which is not the same as equation (6)

So, system E has not the same solution (2,9)

As a conclusion, the systems which have the same solution are:

<u>Options 1 and 3</u>

5 0
2 years ago
Which of the distributions is left skewed?
tino4ka555 [31]

Answer:

i would pick line c and e before i  would do a and e and b fnhwbh bfrhb bdh hejwhjd febhrew

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
The points R(−4, −1) and S(2, −5) are two vertices of right triangle RST. The hypotenuse of the triangle is RS.
Norma-Jean [14]
      If RS is the hypotenuse of the triangle RST and point T is in Quadrant 3, then point T must be the intersection of the lines: x = - 4 and y = - 5.
       Therefore, the coordinates of point T are ( x, y ) = ( - 4, - 5 ) 
       Answer:
       T ( - 4, - 5 )
3 0
3 years ago
Read 2 more answers
Of the cars sold during the month of July, 89 had air conditioning, 99 had an automatic transmission, and 74 had power steering.
DedPeter [7]

Answer:

There are a total of 23 cars with air conditioning and automatic transmission but not power steering

Step-by-step explanation:

Let A be the cars that have Air conditioning, B the cars that have Automatic transmission and C the cars that have pwoer Steering. Lets denote |D| the cardinality of a set D.

Remember that for 2 sets E and F, we have that

|E \cup F| = |E| + |F| - |E \cap F|

Also,

|E| = |E ∩F| + |E∩F^c|

We now alredy the following:

|A| = 89

|B| = 99

|C| = 74

|A \cap B \cap C| = 5

|(A \cup B \cup C)^c| = 24

|A \ (B U C)| = 24 (This is A minus B and C, in other words, cars that only have Air conditioning).

|B \ (AUC)| = 65

|C \ (AUB)| = 26

|B \cap C| = 11

We want to know |(A∩B) \ C|. Lets calculate it by taking the information given and deducting more things

For example:

99 = |B| = |B ∩ C| + |B∩C^c| = 11 + |B∩C^c|

Therefore, |B∩C^c| = 99-11 = 88

And |A ∩ B ∩ C^c| = |B∩C^c| - |B∩C^c∩A^c| = |B∩C^c| - |B \ (AUC)| = 88-65 = 23.

This means that the amount of cars that have both transmission and air conditioning but now power steering is 23.

3 0
2 years ago
Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$
Viefleur [7K]

QUESTION 1

Given that:

f(x)=2x^2+3x-9,

g(x)=5x+11,

and

h(x)=-3x^2+1

Then;

f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)

f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1

Group similar terms;

f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1

Simplify;

f(x)-g(x)+h(x)=-x^2-2x-19

QUESTION 2

Given that;

f(x)=4x-7.

g(x)=(x+1)^2

and

s(x)=f(x)+g(x)

Substitute the functions;

s(x)=4x-7+(x+1)^2

Substitute x=3

s(3)=4(3)-7+(3+1)^2

s(3)=12-7+(4)^2

s(3)=5+16

s(3)=21

QUESTION 3

Given:

f(x)=3x+2

g(x)=x^2-5x-1

f(g(x))=f(x^2-5x-1)

This implies that;

f(g(x))=3(x^2-5x-1)+2

Expand the parenthesis;

f(g(x))=3x^2-15x-3+2

f(g(x))=3x^2-15x-1

QUESTION 4

The given function is;

f(x)=3(x-6)^2+1

Let

y=3(x-6)^2+1

\Rightarrow y-1=3(x-6)^2

\Rightarrow \frac{y-1}{3}=(x-6)^2

\Rightarrow \sqrt{\frac{y-1}{3}}=x-6

\Rightarrow x=6+\sqrt{\frac{y-1}{3}}

The range is:

\frac{y-1}{3}\ge0

y-1\ge0

y\ge1

The interval notation is;

[1,+\infty)

6 0
3 years ago
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