3 X square + 3 X - 36 by 6 x square - 24 x - 30
1 answer:
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Answer:
The answer is
<h2>12.3 cm</h2>Step-by-step explanation:
Since the triangle is a right angled triangle we can use trigonometric ratios to find y
To find y we use cosine
cos∅ = adjacent / hypotenuse
From the question
y is the adjacent
The hypotenuse is 15
So we have
We have the final answer as
<h3>12.3 cm to the nearest tenth</h3>Hope this helps you
To find the z-score for a weight of 196 oz., use
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
Answer:
There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.
Step-by-step explanation:
We notice that the expression on the left of the equation is a quadratic with leading term , which means that its graph is that of a parabola with branches going up.
Therefore, there can be three different situations:
1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.
2) if its vertex is below the x-axis, the parabola's branches are 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 have NO 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 = 3/4:
This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.
We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:
See the graph produced in the attached image.
