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valentina_108 [34]
3 years ago
13

.

Mathematics
2 answers:
shusha [124]3 years ago
7 0

Answer:

-36.

Step-by-step explanation:

First term a1 = 21.

Common difference d = 18 - 21 = -3   (15-18 = -3)

nth term an = a1 + (n - 1)d

So 20th term = 21 + (20-1) -3

= 21 - 3 * 19

= 21 - 57

= -36.

mixas84 [53]3 years ago
4 0

Answer:

-36

Step-by-step explanation:

21

18

15

12

9

6

3

0

-3

-6

-9

-12

-15

-18

-21

-24

-27

-30

-33

-36

please mark brainliest

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docker41 [41]
Here’s the answer hope it helps: x≤−18
4 0
3 years ago
Please answer this question now
Tema [17]

Answer:

Surface Area = 85.75 ft²

Step-by-step explanation:

Surface area of pyramid = ½(Perimeter of triangular base * slant height of pyramid) + Area of triangular base

Perimeter of triangular base = sum of all sides of the triangular base = 5+5+5 = 15 ft

Slant height of pyramid = 10 ft

Area of triangular base = ½*base of triangle*height of triangle = ½*5*4.3 = 10.75 ft²

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Surface Area = ½(15*10) + 10.75

= (15*5) + 10.75

= 75 + 10.75

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8 0
4 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>

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