Answer:
b
Step-by-step explanation:
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IF the original number was an integer, then the digit in the ones place of the final product is a zero ( 0 ). If it wasn't an integer, then any digit can be in the ones place.
K= 13t-2
= 13(2)-2
=26-2
K=24
Using theorem about secant segments we can write,
AB*AH=AG*AC
AC=4,
CG=6
AG=AC+CG=4+6=10
AH=3
AB= AH+HB=AH+x=3+x
(3+x)*3=10*4
9+3x=40
3x=40-9
3x=31
x=31/3≈10.3
HB≈10.3
EG=HB/2 (as radius and diameter)
EG=10.3/2≈5.2
To solve the problem we could separate the figure into three parts. First figure is a triangle, second figure is a rectangle, third figure is a triangle. See image attached.
Solve each area of the figuresFirst figure, a triangle that have 7 units long of the base, and 2 units long of the height.
a = 1/2 × b × h
a = 1/2 × 7 × 2
a = 14/2
a = 7
The area of the first figure is 7 units²
Second figure is a rectangle, the length of the rectangle is 7 units, the width of the rectangle is 4 units.
a = l × w
a = 7 × 4
a = 28
The area of the second figure is 28 units²
Third figure is a triangle, the base is 7 units long and the height is 2 units long.
a = 1/2 × b × h
a = 1/2 × 7 × 2
a = 14/2
a = 7
The area of the third figure is 7 units²
The area of the three figuresarea = first figure area + second figure area + third figure area
area = 7 + 28 + 7
area = 42
The total area of the figures is 42 units²