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If students want to access our math problem solver tool, they need to follow a few simple steps. Simply click on the services tab, then click on the "free academic tool" tab, then choose the "writing tool" option, which will give you to option to choose a math problem solver. Once you have selected the option, the page will give you multiple options, like –
According to your needs, you can choose one option, and our AI-based tool will give you detailed solutions in real time. Here is an example of how our tool works –
M = -2/3
6M + 4 = 0
6M + (4 – 4) = - 4
4-4 = 0 and 6M = - 4
So, after dividing both sides by 6, we get M = - 4/6
If you want to expand it, you can select the options accordingly. For example,
The GCD of 4 and 6 is 2, so -4/6 = -(2x2)/2x3 = 2/2 x -2/3 = -2/3
That means the value of M is -2/3.
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To find the value of x2 + y2, we need to first determine the individual values of x and y.
Given that x + y = 12, we can use this equation to solve for one variable in terms of the other.
Let's solve for x:
x + y = 12
x = 12 - y
Substituting this value of x into the second equation:
xy = 32
(12 - y)y = 32
12y - y2 = 32
y2 - 12y + 32 = 0
We now have a quadratic equation in terms of y. Solving this equation gives us two possible values for y: y = 8 and y = 4.
Substituting these values back into the equation x + y = 12, we can find the corresponding values of x:
For y = 8:
x + 8 = 12
x = 4
For y = 4:
x + 4 = 12
x = 8
Therefore, we have two solutions: (x = 4, y = 8) and (x = 8, y = 4).
Now, we can calculate x2 + y2 for each solution:
For (x = 4, y = 8):
x2 + y2 = 42 + 82 = 16 + 64 = 80
For (x = 8, y = 4):
x2 + y2 = 82 + 42 = 64 + 16 = 80
So, regardless of which solution we choose, the value of x2 + y2 is 80.
To solve the quadratic equation 2 x 2 + x - 300 = 0 using factorization, we need to find two numbers whose product is -600 (the product of the coefficient of x2 and the constant term) and whose sum is the coefficient of x.
The quadratic equation can be factored as follows:
2x2 + x - 300 = 0
First, let's find the factors of -600:
Factors of -600: -1, 1, -2, 2, -3, 3, -4, 4, -5, 5, -6, 6, -8, 8, -10, 10, -12, 12, -15, 15, -20, 20, -25, 25, -30, 30, -40, 40, -50, 50, -60, 60, -75, 75, -100, 100, -150, 150, -200, 200, -300, 300, -600, 600
Now, let's look for two numbers whose sum is 1 (the coefficient of x). The factors above show that the numbers are -24 and 25 since -24 + 25 = 1.
Next, we can split the middle term of the quadratic equation using these two numbers:
2x2 - 24x + 25x - 300 = 0
Now we group the terms:
(2x2 - 24x) + (25x - 300) = 0
Factor out the greatest common factor from each group:
2x(x - 12) + 25(x - 12) = 0
Now, we can see that (x - 12) is a common factor, so we can factor it out:
(x - 12) (2x + 25) = 0
Now we have factored the quadratic equation as a product of two binomials. To find the solutions, we set each factor equal to zero:
x - 12 = 0 or 2x + 25 = 0
Solving for x in each equation gives us the following:
x = 12 or x = -25/2
Therefore, the solutions to the quadratic equation 2x2 + x - 300 = 0 are x = 12 and x = -25/2.
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