How do you find the area above the xaxis, to the left of #x=8#, to the right of #x=5# and below #y=2x+4#?
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To find the area above the xaxis, to the left of ( x = 8 ), to the right of ( x = 5 ), and below ( y = 2x + 4 ), you first need to find the points of intersection between the given lines and then calculate the area of the region bounded by these lines.

Find the points of intersection:
 Set ( y = 2x + 4 ) equal to ( y = 0 ) to find where it intersects the xaxis: ( 0 = 2x + 4 ), which gives ( x = 2 ).
 At ( x = 5 ), ( y = 2(5) + 4 = 14 ).
 At ( x = 8 ), ( y = 2(8) + 4 = 20 ).

Calculate the area of the region:
 The area can be divided into two regions: a triangle and a trapezoid.
 Triangle: base = ( 8  5 = 3 ), height = ( 14  0 = 14 ).
 Trapezoid: bases = ( 20 ) and ( 14 ), height = ( 8  (2) = 10 ).

Calculate the area of each shape:
 Area of triangle = ( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 14 = 21 ).
 Area of trapezoid = ( \frac{1}{2} \times (\text{sum of bases}) \times \text{height} = \frac{1}{2} \times (20 + 14) \times 10 = 170 ).

Add the areas together:
 Total area = ( 21 + 170 = 191 ).
Thus, the area above the xaxis, to the left of ( x = 8 ), to the right of ( x = 5 ), and below ( y = 2x + 4 ) is ( 191 ) square units.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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