PLEASE HELP! THANK YOU!

d1i1m1o1n [39]

Answer:

2 3/4

Step-by-step explanation:

You need to find the greatest common factor of 4 and 2 then you multiply the 2 by 2 to get a 4 and multiply the 1, then you add those together to get 2 3/4.

5 0

2 years ago

Examplelt: A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall

harina [27]

Answer :

4167 bricks.

Explanation :

Since the wall with all its bricks makes up the space occupied by it, we need to find the volume of the wall, which is nothing but a cuboid.

Here,

${\qquad \dashrightarrow{ \sf{Length=10 \: m=1000 \: cm}}}$

$\qquad \dashrightarrow{ \sf{Thickness=24 \: cm}}$

$\qquad \dashrightarrow{ \sf{Height=4 m=400 \: cm}}$

Therefore,

${\qquad \dashrightarrow{ \bf{Volume \: of \: the \: wall = length \times breadth \times height}}}$

${\qquad \dashrightarrow{ \sf{Volume \: of \: the \: wall = 1000 \times 24 \times 400 \: {cm}^{3} }}}$

Now, each brick is a cuboid with Length = 24 cm, Breadth = 12 cm and height = 8 cm.

So,

${\qquad \dashrightarrow{ \bf{Volume \: of \: each \: brick = length \times breadth \times height}}}$

${\qquad \dashrightarrow{ \sf{Volume \: of \: each \: brick = 24 \times 12 \times 8 \: {cm}^{3} }}}$

So,

${\qquad \dashrightarrow{ \bf{Volume \: of \: bricks \: required = \dfrac{volume \: of \: the \: wall}{volume \: of \: each \: brick} }}}$

${\qquad \dashrightarrow{ \sf{Volume \: of \: bricks \: required = \dfrac{1000 \times 24 \times 400}{24 \times 13 \times 8} }}}$

${\qquad \dashrightarrow{ \sf{Volume \: of \: bricks \: required = \bf \: 4166.6} }}$

Therefore,

<u>The wall requires 4167 bricks. </u>

6 0

2 years ago

Find the surface area of each figure pls 7-10

Vlad [161]

Answer:

7) 42 $inches^{2}$

8) 358 $meters^{2}$

9) 320 $yards^{2}$

10) 288 $miles^{2}$

Step-by-step explanation:

Problems 7, 8, and 10 all use the same formula: the formula for the surface area of a rectangular prism.

Formula: 2 * (lw + lh + hw)

where l = length, w = width and h = height

This is a condensed form of the sum lw + lw + lh + lh + hw + hw where each product is a face of the prism.

replace each variable in the formula above with the values given in your problems for 7, 8 and 10:

7) 2 * (3*3 + 3*2 + 2*3) = 2 * (9 + 6 + 6) = 2 (21) = 42

8) 2 * (12*7 +12*5 + 5*7) = 2 * (84 + 60 + 35) = 2(179) = 358

10) 2* (12*4 + 12*6 + 4*6) = 2 * (48 + 72 + 24) = 2(144) = 288

Now 9 is a little different, as it contains a regular pentagon. However it is not much different from the other problems as it is also a prism.

The formula for the area of any regular polygon can be in the form:

Area = $\frac{1}{2}$ ap where the a stands for apothem (which is the distance from the center of a regular polygon to its side, in this case it is 5.5 yds) and the p is the perimeter of the polygon.

The reason for this formula involves the polygon being divided into as many congruent triangles as it has sides, but that is for another time : )

Given this formula for the area of the pentagon, you can substitute values:

Area = $\frac{1}{2}$ (5.5)(number of sides * length of a side)

Area = $\frac{1}{2}$ (5.5)(5 * 8) = 110

And get that the area of the face is 110 yd^2

the rest of the faces are rather straightforward as they contain rectangles of sides 4 yds and 5 yds, and you would end up with 5 of them (for each side of the pentagon)

Therefore by adding up each of the bases and the sides, you should get:

110 + 110 + 20 + 20 + 20 + 20 + 20 = 320

Hope that helped : )

3 0

3 years ago

The answer must be a mixed number a proper fraction or a integer

mina [271]

Answer:

i dont understand sir sorry

8 0

2 years ago

For a circle of raduis 6feet, find the arc length of a central angle of 6

murzikaleks [220]

Length of a circumference=2πr
r=radius=6 feet.
lengt of this circumference=2π6 feet=12π feet.

1 circumference=360º; then:

360º-------------------12π feet
6º---------------------- x

360º * x=6º * 12π feet
x=(6º * 12π feet) / 360º=π/30 feet (≈0,105 feet).

Solution: π/30 feet

4 0

3 years ago