Answer:
≈ 0.9 cm²
Step-by-step explanation:
I don't know if there is any easier way to solve this, I use 8th grade slope and line method to solve. suppose A is on origin, then the coordinates of triangle ABC are (0,0) (1,3) (4,1) and polygon DEFGH intersect line BC at F and G.
F is the intersect of y=2 and line BC, while G is intersect of x=3 and line BC.
Line BC: pass (1,3) and slope (m) = (1-3)/(4-1) = - 2/3,
F(x,2)
(2-3) / (x-1) = -2/3
2x-2 = 3 x = 2.5
*<u>F (2.5 , 2</u>)
G (3,y)
(y-3) / (3-1) = -2/3 y = 5/3
*<u>G (3 , 5/3)</u>
*<u>point I (3 , 2</u>)
segment FI = 3-2.5 = 0.5 = 1/2
segment GI = 2 - 5/3 = 1/3
area ΔFIG = (1/2 * 1/3) / 2 = 1/12 = 0.083
area of polygon DEFGH (region colored RED) = 1 - 0.083 = 0.917 ≈ 0.9 cm²
X^(5/8)/x(1/4)= x^(5/8)/x^(2/8)= x^(3/8)
Answer:
3.55-3.59
Step-by-step explanation:
to round off to the nearest whole number, look at the first number after the decimal, if it is less than 5, add zero to the units term, If it is equal or greater than 5, add 1 to the units term.
So numbers that would be rounded off to 3.6 have to have an hundredth value equal or greater than 5
they include
3.55
3.56
3.57
3.58
3.59
Step-by-step explanation:
f is (x+1)-1
and g (the next letter in the alphabet) is x-2
first for f
x+1-1 = x
Then we -2 for g
So every letter there o words is just 2 less than the previous letter like
h would be h= x-4
so r(x) is g which is x-2 and x which would have to be the remaining amount to get to the number
Answers: perpendicular bisector; image
- The reflection of a point P across a line m is the point P' if line m is the <u> perpendicular bisector </u> of segment PP'
- Point P' is called the <u> image </u> of point P.
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Explanation:
If we were to reflect point P over line m to land on P', then we've gone from pre-image to image.
Let's say that Q is the point on segment PP' and it's also on line m. We consider Q the midpoint of segment PP' which means PQ and P'Q are the same length. This indicates P and P' are the same distance away from the mirror line, and that point Q bisects segment PP'. Also, line m is perpendicular to segment PP'. So that's where the "perpendicular bisector" comes from.