The square root of 192 is 13.856. so the answer is more than likely 14 or 13.85
Answer:
2.7 in²
Step-by-step explanation:
similar triangles have the same angles, and all side lengths (or other distances) of one triangle have the same scaling factor to the side lengths of the other triangle.
so, we know the relation between the 2 baselines is 2/3, as this is the factor to turn the baseline of the large triangle into the baseline of the smaller triangle.
in other words
EF = BC × 2/3
2 = 3 × 2/3
correct
how do we calculate the area of a triangle ?
Area = baseline × height / 2
from BAC we know
Area = 6
baseline = 3
height = ?
6 = 3 × height / 2
12 = 3 × height
height = 4
aha !
now, EDF has a height too that we need to calculate is Area. and this height has the same scaling factor compared to the larger triangle as the side lengths : 2/3
so, for EDF we know
Area = ?
baseline = 2
height = 4 × 2/3 = 8/3
therefore, the area is
Area = (2 × 8/3) / 2 = (16/3) / 2 = 8/3 = 2.66666... ≈ 2.7
the shirt answer would be :
we know from the 2 baselines that the scaling factor for each distance is 2/3.
for the area we need to multiply 2 distances, so that means we have to multiply both by 2/3. and so on the formula for the area we have to use 2/3 × 2/3.
2/3 × 2/3 = 4/9
=>
Area small = Area large × 4/9 = 6 × 4/9 = 24/9 = 8/3 ≈ 2.7
Answer:
- 60<em><u>÷</u></em><em><u>15</u></em><em><u> </u></em><em><u>=</u></em><em><u>4</u></em><em><u> </u></em>
<em><u>Therefore</u></em><em><u> </u></em><em><u>Martin</u></em><em><u> </u></em><em><u>uses</u></em><em><u> </u></em><em><u>his</u></em><em><u> </u></em><em><u>power</u></em><em><u> </u></em><em><u>saw</u></em><em><u> </u></em><em><u>4</u></em><em><u> </u></em><em><u>times</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>hour</u></em><em><u> </u></em>
Answer:
J
-66
Step-by-step explanation:
a1 = 2
n = 18
d = - 4
a18 = a1 + (n - 1) * d
a18 = 2 + (18 - 1)*(-4)
a18 = 2 + 17*-4
a18 = 2 - 68
a18 = -66
<u>Answer:</u>
<u>Step-by-step explanation:</u>
We have a parabola and a small line shown in the graph.
The parabola goes upto x = 2 but does not reach that very value. So x = 2 starts on the line above where x = 2 and ends just before x = 4.
Therefore, this function can be modeled by:
