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-BARSIC- [3]
2 years ago
8

Find the area of the trapezoid below

Mathematics
1 answer:
ki77a [65]2 years ago
5 0
The total are is 828 cm
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artcher [175]

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X= -0.75

Step-by-step explanation:

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In your lab, a substance's temperature has been observed to follow the function T(x) = (x + 5)3 + 7. The turning point of the gr
OleMash [197]
So essentially, this is a cubic function where T(x) = y, and:
y =  {(x+5)}^{3} +7
In order to find the vertices, aka "turning points", we need to see around what area the graph will curve quickly from bottom-left to top-right. If the value being cubed (in the parentheses) is around -8, y will be close to 0, since -8+7=-1. (-2)^3 = -8. So x+5=-2, x=-7.
At x=-6, y = (-6+5)^3+7 = (-1)^3+7
y = -1+7 = 6
At x=-5, y = 0+7 = 7
Above x=-4, the y jumps rapidly above 10, and below x=-9, the y jumps rapidly below -50
What does all that mean???
That between x = -9 and x = -4 (x = -4,-5,-6,-7,-8,-9), the temperature (T(x) or y) does not change very much. Therefore, around those temperatures is a turning point, where the phase shift occurs from solid/liquid.
4 0
3 years ago
While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at
tekilochka [14]

The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.

<h3>How to estimate the speed of the moving walkway relative to the airport terminal?</h3>

Let x be the speed of the walkway.

(2.8 + x) = speed of child moving in direction of the walkway

(2.8 - x) =  speed of child moving against the direction of the walkway

Travel time = distance/speed

Travel time of child moving in direction of walkway = 23/(2.8+x)

Total elapsed time given = 29s

23/(2.8 + x)+ 23 / (2.8-x) = 29

LCD = (2.8 + x)(2.8 - x)

23(2.8 - x) + 23(2.8 + x) = 29(2.8 + x)(2.8 -x)

simplifying the equation, we get

23*2.8-23x+23*2.8+23x=29(2.8^2-x^2)

23(2.8+2.8)/29=2.8^2-x^2

x^2=(2.8)^2-(23*5.6)/29)=3.4

x=\sqrt{3.4}=1.84m/s

Speed of walkway = 1.84 m/s

The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.

To learn more about Speed refer to:

brainly.com/question/4931057

#SPJ4

4 0
1 year ago
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