9514 1404 393
Answer:
3433 m^2
Step-by-step explanation:
The total surface area is the sum of the end areas and the lateral area.
The end areas together are a circle of diameter 0.4 m and two rectangles that are 0.4 m by 0.55 m.
End areas = πr^2 + 2LW = 3.14(0.2 m)^2 + 2(0.4 m)(0.55 m)
= 0.1256 m^2 +0.44 m^2 = 0.5656 m^2
The perimeter of the end piece is half the circumference of a 0.4 m circle and three sides of a 0.4 m by 0.55 m rectangle.
P = (1/2)πd + 2L +W = 1/2·3.14·0.4 m + 2(0.55 m) + 0.4 m = 2.128 m
The lateral area of the mailbox is the product of its length and the perimeter of the end piece:
LA = PL = (2.128 m)(0.6 m) = 1.2768 m^2
Then the total surface area of the mailbox is ...
0.5656 m^2 + 1.2768 m^2 = 1.8424 m^2
__
The area of aluminum required for 1863 mailboxes is ...
1863 × 1.8424 m^2 ≈ 3433 m^2
About 3433 square meters of aluminum are needed to make these mailboxes.
i would say B
when it says "the measure", it's refering to where the second quartile is beginning
Answer:
d. 18
Step-by-step explanation:
3x - 5 = 22
3x = 27
x = 9
Then, 2x is equal to:
= 2(9)
= 18
<span>So we want to know how much ball bearings can be made with 5.24 cm^3 if one ball bearing has a diameter of 1 cm. We know that radius r=d/2=0.5cm So the volume V of one ball bearing is: V=(4/3)*pi*r^3 so V=(4/3)*3.14*(0.5)^3cm^3=0.524cm^3. Now we simply divide the volume of steel by the volume of the ball bearing: 5.24/0.524=10. So we can make 10 ball bearings from 5.24 cm^3 of steel</span>
1/8 into a decimal is 0.125.
