The real solution occurs when the graph intersects the x axis,
In the problem shown, the graph does not intersect the x axis, therefore it has no real solution, this means that the answer must have a complex conjugate pair
The answer is
function f has exactly two complex solutions
Answer:
A
Step-by-step explanation:
<span>The <u>correct answer</u> is:
$31.40.
Explanation<span>:
Since 6 oranges cost $3.00, we can divide to find the cost of one orange:
3.00/6 = $0.50. Each orange costs $0.50.
If ten oranges are bought, this would cost 10(0.50)=$5.00.
Each grapefruit costs $2.40; 11 of them would cost 11(2.40) = 26.40.
Together the total cost is $5+$26.40 = $31.40.</span></span>
Answer:
51 m^2
Step-by-step explanation:
The shaded area is the difference between the area of the overall figure and that of the rectangular cutout.
The applicable formulas are ...
area of a triangle:
A = (1/2)bh
area of a rectangle:
A = bh
area of a trapezoid:
A = (1/2)(b1 +b2)h
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We note that the area of a triangle depends only on the length of its base and its height. The actual shape does not matter. Thus, we can shift the peak of the triangular portion of the shape (that portion above the top horizontal line) so that it lines up with one vertical side or the other of the figure. That makes the overall shape a trapezoid with bases 16 m and 10 m. The area of that trapezoid is then ...
A = (1/2)(16 m + 10 m)(5 m) = 65 m^2
The area of the white internal rectangle is ...
A = (2 m)(7 m) = 14 m^2
So, the shaded area is the difference:
65 m^2 -14 m^2 = 51 m^2 . . . . shaded area of the composite figure
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<em>Alternate approach</em>
Of course, you can also figure the area by adding the area of the triangular "roof" to the area of the larger rectangle, then subtracting the area of the smaller rectangle. Using the above formulas, that approach gives ...
(1/2)(5 m)(16 m - 10 m) + (5 m)(10 m) - (2 m)(7 m) = 15 m^2 + 50 m^2 -14 m^2
= 51 m^2
The outlier (61) is at the low end of the data set, but doesn't affect the mean by a lot, so ...
The mean is centered among the other numbers in both sets of data.
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The mean without the outlier is 114. With the outlier, it is 107.4. The lower quartile is 108, so the mean does get moved outside the "box" of the box-and-whisker plot of the data set without the outlier.
