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2 years ago

Need help help me guys,

Mathematics

1 answer:

Ganezh [65]2 years ago

8 0

Hello!

$\large\boxed{x + 2}$

$\frac{2x^{2} +x}{2x-2} + \frac{x-4}{2x-2}$

The denominators are equal, so we can simply combine the numerator:

$\frac{2x^{2} +x+ x - 4}{2x-2}$

Combine like terms in the numerator:

$\frac{2x^{2} +2x - 4}{2x-2}$

Factor out 2 from the entire fraction:

$\frac{2(x^{2} +x - 2)}{2(x-1)}$

Cancel out 2 from both the numerator and denominator:

$\frac{x^{2} +x - 2}{x-1}$

Factor the numerator to simplify further:

$\frac{(x + 2)(x - 1)}{x-1}$

Cancel out x - 1:

$x + 2$

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The sum of 6 consecutive integers is 204. what is the fourth number in this sequence?

zloy xaker [14]

Let $x$ by the smallest of the 6 integers. Then

$x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=204$

$\implies6x+15=204$

$\implies6x=189$

But 189 is odd, so it cannot be divisible by 6, which means $x$ is not an integer...

8 0

3 years ago

Find the next two terms of the following sequence: 14,38,74,122,182,254

Roman55 [17]

ANSWER

The next two terms are

338, 434

EXPLANATION

The given sequence is 14,38,74,122,182,254

Let us observe some pattern and use it to find the next two terms.

14+24=38

38+36=74

74+48=122

122+60=182

182+72=254

To get the next term we add 84 to 254

254+84=338

To get the next term,we add 96 to 338

338+96=434

8 0

3 years ago

JUST ANSWER PLEASE!!! QUICK

jeka57 [31]

Answer:

Options 1 and 3

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.

System A) 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)

System B) 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)

System C) (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)

System D) (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) OR (T-P) and (Q-U)

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

System E) (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:

Options 1 and 3

5 0

2 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

3 years ago

In this sum, each letter stands for one of the digits 0 to 9

lora16 [44]

Answer:

7 4 5 0

F O U R

4 8 1

O N E

=7 9 3 1

F I V E

Hope this is correct:)

6 0

2 years ago

Need help help me guys,​

Need help help me guys,