12 4/5 can be changed as 64/5 so there are 64 1/5ths
Well the answer would be 17 .
Answer: (1,-1)
Step-by-step explanation:
Midpoint of BC=(6+4)/2, (3–1)/2. =(5,1)
Slope of BC is (3+1)/4–6)= 4/-2 = -2
Slope of perpendicular bisector of BC =+1/2
Eqn of perpendicular bisector is : Y-1 =1/2 (x-5)
Y=1/2 •(x-5) +1
Midpoint of AB. (6–2)/2, (3–1)/2 ={2,1)
Slope of AB is(3+1)/(-2–6) = 4/-8 =-1/2
Slope of perpendicular bisector = +2
Eqn of perpendicular bisector is Y-1. =2( X-2)
Y=2X-4+1 = 2X -3
Solving Y=(X-5)/2 +1
& Y=2X-3
2X-3 =(x-5/2)+1
2X-4 =(x-5/2)
4X-8 = x-5
3X =3
X=1
Y= 2×1–3= -1
Circumcentre is(1,-1)
For the first question break it down to 2 rectangles and add the areas
14 * 4
and (12-4) * 8 = 8* 8
secon q:-
The other rectangle which has area of 72 is the third choice. Its perimmeter = 2(24) + 2(8) = 64
<u><em>Answer:</em></u>
<u><em>h(t) = t^2=5+ for = -2,0,5,7</em></u>
<u><em></em></u>
<u><em>1. h(t)=t^2=5= -2 Only = -2</em></u>
<u><em>2. h(t)=t^2 = 5 = -5 Only = -5</em></u>
<u><em>3. h(t) = t^2 = 5 = -5 only = -5</em></u>
<u><em>4. h(t) = t^2 = 5 = -7 Only = -7</em></u>
<u><em></em></u>
<u><em>1.h(t) = t^2 = 5 = 2 </em></u>
<u><em>Slope: 0</em></u>
<u><em>y - intercept: (0,2)</em></u>
<u><em></em></u>
<u><em>2.h(t) = t^2 = 5 = 0</em></u>
<u><em>Slope: 0</em></u>
<u><em>y- intercept: (0,0)</em></u>
<u><em></em></u>
<u><em>3.h(t) = t = 5 =5</em></u>
<u><em>Slope: 0</em></u>
<u><em>y - intercept (0.5)</em></u>
<u><em></em></u>
<u><em>4.h(t) = t^2 = 5 = 7</em></u>
<u><em>Slope: 0</em></u>
<u><em>y - intercept:(0,7)</em></u>
<u><em></em></u>
<u><em>Hope this helps:3</em></u>
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