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guajiro [1.7K]
3 years ago
11

What is the sum of 56+64 writen as the product of its gcf and another sum

Mathematics
1 answer:
jenyasd209 [6]3 years ago
3 0
<span>Look for the sum of 56 and 64, written as the product of its GCF and another sum.
First, let’s find the greatest common factor of both given numbers:
=> 56 = 1, 2, 4, 7, 8, 13, 28 and 56
=> 64 = 1, 2, 4, 8, 16, 32, and 64
Now, we need to find the greatest common factor between the two numbers. The GCF of the 2 numbers is 8.
=> 56 / 8 = 7
=> 64 / 8 = 8
=> 56 + 64 = 120
=> (8 x 7) + (8 x 8)
=> 56 + 64
=> 120.</span><span>

</span>



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Step-by-step explanation:

200+(20-4)-2x3

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Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
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The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term x^2, which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form f(x)=ax^2+bx+c is given by the expression: x_v=\frac{-b}{2a}

Since in our case a=1 and b=5, we get that the x-position of the vertex is: x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the x_v=-\frac{5}{2}. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

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