1. The shape of cross-section is a circle.
2. The face parallel to ABCD is EFGH. Since this is a a rectangular shape,
A = L*H = 12*6 = 72 cm^2
3. The cross-section parallel to ABC is DEF with h = 12 ft, b= 5ft (where h is the height and b is the base of a right angled triangle).
Area, A = 1/2 *b*h = 1/2*5*12 =30 ft^2
4. Plane BDHF is a rectangle shape whose length is the diagonal of ABCD.
Diagonal BD = sqrt (AB^2+BD^2) = sqrt (8^2+7^2) = 10.63 cm.
Perimeter, P = 2(BD+DH) = 2(10.63+6) = 33.26 cm
<span>First we have to determine the slope of each lines by transforming to the slope-intercept form:
y=(3x-7/)4; m2= ¾y=(12x+6)/5, m3 = 12/5
The formula to be used in the proceeding steps is a=tan^-1(m1-m2)/1+m1m2=tan^-1(m1-m2)/1+m1m2
substituting, a=tan^-1(m1-3/4)/1+3m1/4=tan^-1(m1-12/5)1+12m1/5) =>(4m1-3)/(4+3m1)=(5m1-12)/(5+12m1)m1 = -1applying this slope
y -y1 = m(x-x1)
when y1 = 5 and x1 = 4 then,
y - 5 = -1(x-4)
y = -x +4+ 5 ; y = -x +9</span>
48 i believe thats the best i can find
You could directly add 6+4 which gives you the answer as well... M=10
Answer:
the answer is 9730.06
Step-by-step explanation:
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