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3 years ago

2x-4=-4x+6 What value of x makes the equation true A x=5/3 B x=-5/3 C x=5 D x=-5

Mathematics

2 answers:

Rama09 [41]3 years ago

8 0

Answer: 5/3

6x = 10

x=10/6

x=5/3

Arte-miy333 [17]3 years ago

6 0

Answer:

x = 5/3

Step-by-step explanation:

2x-4=-4x+6

Add 4x to each side

2x+4x-4=-4x+4x+6

6x -4 = 6

Add 4 to each side

6x -4 +4 = 6+4

6x = 10

Divide each side by 6

6x/6 = 10/6

x = 5/3

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kap26 [50]

Answer:

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Step-by-step explanation:

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3 years ago

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

AysviL [449]

Answer:

a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution

6 0

3 years ago

Mr. Hammond is the volunteer coordinator for a company that puts on running races. Last year, the company organized 7 short race

worty [1.4K]

Answer:

28 short race volunteers

76 long race volunteers

Step-by-step explanation:

Let :

Short races = a

Long races = b

LAST YEAR:

7a + 8b = 804 - - - - (1)

THIS YEAR:

11a + 5b = 688 - - - (2)

Multiply (1) by 5 and (2) by 8

35a + 40b = 4020

88a + 40b = 5504

Subtract

-53a = - 1484

a = 1484 / 53

a = 28

Put a = 28 in equation (1)

7(28) + 8b = 804

196 + 8b = 804

8b = 804 - 196

8b = 608

b = 76

28 short race volunteers

76 long race volunteers

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2 years ago

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lora16 [44]

Answer:

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Step-by-step explanation:

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3 years ago

A bag contains 3 green marbles, 7 blue marbles and 2 black marbles. The probability of randomly picking a green marble is 1/4. W

Dima020 [189]

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3 years ago