Answer:
169.5 yd²
Step-by-step explanation:
See attachment.
In situations like this, ALWAYS (!) make as sketch.
Divide the areas by adding dotted lines on sensible places, and write in the missing numbers for the correct distances.
The only possible difficult one is the triangle, which is the half of a rectangle. So you are dealing with a series of areas of rectangles. That is really easy if you understand what you are doing.
Total area =
area 1 + area 2 + area 3 + area 4
10*4 + 4*7 + 3*13 + (0.5 * 7*17)
40 + 28 + 42 + 59.5
Total area = 169.5 yd²
The surface area is ≈ 451.92
The equation that works for this is
y=1/7x + 0! It is parallel and goes through (0,0)
Hope this helps :)
Answer:
idk
Step-by-step explanation:
idk just responding
for points
Answer:
a) maximum; the parabola opens downward
b) positive; it must lie above the x-axis
c) x = 1.5
Step-by-step explanation:
The x-intercepts of a function are the points where the graph of the function crosses the x-axis. The y-values there are zero.
The "differences" of a function are related to the average slope between adjacent points. Second differences are related to the rate of change of the slope of the function. When <em>second differences are negative</em>, as here, the slope of the quadratic function is decreasing, becoming more negative. We say the <em>curvature</em> of the function is <em>negatve</em>, and that it <em>opens downward</em>.
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<h3>a, b.</h3>
If the graph of the parabola opens downward, and it crosses the x-axis, it must have a <em>maximum</em> that is a <em>positive value of y</em>.
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<h3>c.</h3>
The graph of a parabola is symmetrical about its vertex. That means points on the same horizontal line are the same distance from the line of symmetry, which must go through the vertex. The x-coordinate of the vertex will be the x-coordinate of the midpoint between the two x-intercepts:
x = (-2 +5)/2 = 3/2
The x-coordinate of the vertex is x = 1.5.
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<em>Additional comment</em>
The attachment shows a table with three evenly-spaced points on the curve. The calculations show first differences (d1) and second differences (d2). You can see that the sign of the second diffference is negative, in agreement with the given conditions.
