Hello!
To find the surface area of a cube you do
where "a" is the length of one of the sides.
So
= 1350
the answer is 1350mm
Hope this helps!
To find the surface area of a cube you do
So
the answer is 1350mm
Hope this helps!
Factor: x² – 3x+2 O A. (3-2)(x-1) O B. (x-2)(x+1) OC. (x+2)(x+1) OD. (x+2)(x-1)
1 answer:
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Answer:
The function h(x) is decreasing on the interval (3, ∞).
Step-by-step explanation:
Please take a look at the attached image.
You will see a graph of the given function h(x) = -2. The function is decreasing.
The function starts at 3 and starts to go towards negative infinity on the x-axis. Therefore the function is decreasing on the interval (3, ∞).
Answer:
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 , 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 is given by the expression:
Since in our case and
, we get that the x-position of the vertex is:
Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:
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 . 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):
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.
