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andreev551 [17]
2 years ago
14

HELP PLEASE I'M WRITING A TEST!!! x+3/4 = x+5/6

Mathematics
1 answer:
Keith_Richards [23]2 years ago
5 0

Answer:

no solution

Step-by-step explanation:

if u plug a random number like 1 for example 1+3/4=1 3/4 and 1+5/6=1 5/6

1 3/4 =1 5/6-this is NOT equal

so no solution

hopefully that helps! please let me know if i did anything wrong thanks!

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Step-by-step explanation:

According to the associative property of addition, you can put the parentheses anywhere you like. This is the only one that works.

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Find the zeros of the function. f(x) = x2 - 6x + 8
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The zeroes of this function are x = 2, 4.

We can find this by factoring. Factoring x²-6x+8, we get (x-2)(x-4). Now, since we want to find the zeroes, we have to make y equal to zero, or (x-2)(x-4) = 0. Using the zero-product property, we can conclude that if (x-2)(x-4) is 0, x is 2, 4. 
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Give a geometric description of the following system of equations.a. 2x−4y=12 −3x+6y=−15.b. 2x−4y=12 −5x+3y=10.a. 2x−4y=12 −3x+6
Reptile [31]

Answer:

a. No solution, parallel lines.

b. One solution.

Step-by-step explanation:

Given the system of equations:

a. 2x-4y=12

-3x+6y=-15

b. 2x-4y=12

-5x+3y=10

To give a geometric description of the given system of equations.

The geometric description of a system of equations in 2 variables mean the system of equations will represent the number of lines equal to the number of equations in the system given.

i.e.

Number of planes = Number of variables

Number of lines = Number of equations in the system.

Here, we are given 2 variables and 2 equation in each system.

So, they can be represented in the xy-coordinates plane.

And the number of solutions to the system depends on the following condition.

Let the system of equations be:

A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0

1. One solution:

There will be one solution to the system of equations,  If we have:

\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}

2. Infinitely Many Solutions: (Identical lines in the system)

\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}= \dfrac{C_1}{C_2}

3. No Solution:(Parallel lines)

\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}

Now, let us discuss the system of equations one by one:

a. 2x-4y=12 OR 2x-4y-12=0

-3x+6y=-15 OR -3x+6y+15=0

A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -3, B_2 = 6, C_2= 15

Here, the ratio:

\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2} = -\dfrac{2}{3}\\\dfrac{C_1}{C_2} = -\dfrac{4}{5}

\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}

Therefore, no solution i.e. parallel lines.

b. 2x-4y=12 OR 2x-4y-12=0

-5x+3y=10 OR -5x+3y-10=0

A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -5, B_2 = 3, C_2 = -10

\dfrac{A_1}{A_2}= -\dfrac{2}{5}\\\dfrac{B_1}{B_2} = -\dfrac{4}{3}\\\dfrac{C_1}{C_2} = -\dfrac{6}{5}

\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}

So, one solution.

Kindly refer to the images attached for the graphical representation of the given system of equations.

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