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ohaa [14]
3 years ago
11

i am a solid figure with two congruent polygons that are bases,connected with lateral face that are rectangles.

Mathematics
1 answer:
BabaBlast [244]3 years ago
6 0
Rectangular prisim is the answer
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Can someone help me with this?
uysha [10]

Step-by-step explanation:

Volume of a cylinder : πr^2*height.

Volume : 150cm^3

height : 7.2m

π =3.14

150 = 3.14*r*r*7.2

150=22.608*r*r

r^2= 150/22.608

=150/23cm^2

r=√6.5

r=2.569cm

r=2.6cm

8 0
3 years ago
Simplify the expression 56-2(4+x)
lesya692 [45]

Answer:

48+x

Step-by-step explanation:

56-2(4+x)

56-8+x

48+x

5 0
3 years ago
Where does the graph of y = sin x from x = 0 to x = 2π start?
Bad White [126]

the graph starts at an x intercept.

6 0
3 years ago
Read 2 more answers
QUESTION :-
Galina-37 [17]

Answer:

  • 92000

Step-by-step explanation:

<u>Given:</u>

  • I = 5% = 0.05
  • t = 2 years
  • Let the sum is x.

<u>Simple interest:</u>

  • SI = x*2*0.05 = 0.1x

<u>Compound interest:</u>

  • CI = x*(1 + 0.05)² - x = 1.1025x - x = 0.1025x

<u>The difference is 230:</u>

  • 0.1025x - 0.1x = 230
  • 0.0025x = 230
  • x = 230/0.0025
  • x = 92000
3 0
3 years ago
(2*)2-3×2*+2=0<br>4m-15°(×)m+75°​
Valentin [98]

Answer:

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

7 0
2 years ago
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