Answer:
180 sq.cm
Step-by-step explanation:
A = 10 x 10 + 16 x 5 x 1/2 x 2
A = 100 + 16 x 5
A = 100 + 80
A = 180 square cm
Hope that helps!
Answer:
Step-by-step explanation:
If you graph there would be two different regions. The first one would be

And the second one would be
.
If you rotate the first region around the "y" axis you get that

And if you rotate the second region around the "y" axis you get that

And the sum would be 2.51+4.188 = 6.698
If you revolve just the outer curve you get
If you rotate the first region around the x axis you get that

And if you rotate the second region around the x axis you get that

And the sum would be 1.5708+1.0472 = 2.618
X^2-11x+28=0
(x-7)(x-4)=0
x-7=0
x=7
x-4=0
x=4
Answer:
1/6
Step-by-step explanation:
Let x represent the proportion of time the third person spends on the project. You want ...
1/2 + 1/3 + x = 1 . . . . full-time equivalents
5/6 + x = 1 . . . . simplify
x = 1/6 . . . . . . . subtract 5/6
The third person should budget 1/6 of their time to the project.
The required proof is given in the table below:
![\begin{tabular}{|p{4cm}|p{6cm}|} Statement & Reason \\ [1ex] 1. $\overline{BD}$ bisects $\angle ABC$ & 1. Given \\ 2. \angle DBC\cong\angle ABD & 2. De(finition of angle bisector \\ 3. $\overline{AE}$||$\overline{BD}$ & 3. Given \\ 4. \angle AEB\cong\angle DBC & 4. Corresponding angles \\ 5. \angle AEB\cong\angle ABD & 5. Transitive property of equality \\ 6. \angle ABD\cong\angle BAE & 6. Alternate angles \end{tabular}](https://tex.z-dn.net/?f=%20%5Cbegin%7Btabular%7D%7B%7Cp%7B4cm%7D%7Cp%7B6cm%7D%7C%7D%20%0A%20Statement%20%26%20Reason%20%5C%5C%20%5B1ex%5D%20%0A1.%20%24%5Coverline%7BBD%7D%24%20bisects%20%24%5Cangle%20ABC%24%20%26%201.%20Given%20%5C%5C%0A2.%20%5Cangle%20DBC%5Ccong%5Cangle%20ABD%20%26%202.%20De%28finition%20of%20angle%20bisector%20%5C%5C%20%0A3.%20%24%5Coverline%7BAE%7D%24%7C%7C%24%5Coverline%7BBD%7D%24%20%26%203.%20Given%20%5C%5C%20%0A4.%20%5Cangle%20AEB%5Ccong%5Cangle%20DBC%20%26%204.%20Corresponding%20angles%20%5C%5C%0A5.%20%5Cangle%20AEB%5Ccong%5Cangle%20ABD%20%26%205.%20Transitive%20property%20of%20equality%20%5C%5C%20%0A6.%20%5Cangle%20ABD%5Ccong%5Cangle%20BAE%20%26%206.%20Alternate%20angles%0A%5Cend%7Btabular%7D)