Answer:
6.25 % of area of Large triangle = area of smallest triangle
Step-by-step explanation:
Area of large triangle L = (1/2)*base*height
L = (1/2)*(44 m)* h
L = 22*h square m.
2nd equilateral triangle area: S =(1/2)*(22 m)*(0.5h)
S = 5.5*h sq. m.
3rd smallest equilateral triangle area : T = (1/2)*(11 m)*(0.25h)
T = 1.375*h sq. m
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Find percent P where T = P* L, 1.375*h = P * 22*h
P = 6.25% = 0.0625
Answer:
∅1=15°,∅2=75°,∅3=105°,∅4=165°,∅5=195°,∅6=255°,∅7=285°,
∅8=345°
Step-by-step explanation:
Data
r = 8 sin(2θ), r = 4 and r=4
iqualiting; 8.sin(2∅)=4; sin(2∅)=1/2, 2∅=asin(1/2), 2∅=30°, ∅=15°
according the graph 2, the cut points are:
I quadrant:
0+15° = 15°
90°-15°=75°
II quadrant:
90°+15°=105°
180°-15°=165°
III quadrant:
180°+15°=195°
270°-15°=255°
IV quadrant:
270°+15°=285°
360°-15°=345°
No intersection whit the pole (0)
Slope = (9-3)/(5-2) = 6/3 = 2
answer
<span>A) 2 </span>
We can find this integer using a system of equations. The first thing we look at is the first sentence.
Let's call X The first integer,
Let's call Y the second integer.
So, X+Y=61 and X-Y= 1
Using Substituion on the second equation, we get
X=Y+1
(Now we plug in Y+1 into the second equation for X)
So,
(Y+1)+Y=61
2Y=60
Y=30
So, one of the integers is 30.
Now to find the second, we go back to
X=Y+1
X=30+1
X=31.
So, the integers are 31 and 30.
Hope this helps!
