A rectangle has a length of 14 meters and a width of 200 centimeters what is the area of the rectangle in meters?

Answers

Answer 1
Answer:

Answer:

28 m

Step-by-step explanation:

1 m=100 cm

?=200 cm ...

200 cm=2m

area=lw

A=14 m*2 m

A=28 m

Answer 2
Answer:

Answer:

28

Step-by-step explanation:

200 cm = 2m

Area = length × Width

So..

Area = 14×2

= 28


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Brainliest if correct hurryyy!!

Which of the following equations has exactly one real solution?A.
2(x + 14) = 2x + 28
B.
7x + 14 = −2x − 17
C.
2x + 14 = 2x + 14
D.
7x + 14 = 7x − 17

Answers

Answer:

Option A

Step-by-step explanation:

2(x+14)=2x+28

2x+28=2x+28

IT HAS REAL SOLUTION :)

PLZZ MARK ME AS BRAINLIEST

Guys, can you please help me find the 2 values of k so that an each trinomial can be factored over the integers for me, please? :) a). 2x²–7x+k
b). kx²–5x+3

Answers

Answer:

~Senpi Boi here~

Step-by-step explanation:

For the values of a and b that can be factored are the following answers, for a it would be 2x^2-6x. And for b it would be x-5/2^2.

(Hope this helps!) ^-^

WXYZ IS A RECTANGLE. What is ZX?

Answers

ZX is one side out of the triangle 
ZX is a diagonal line from one point (corner) of the rectangle to the the other point (corner).

Hoped I helped!

the area of the rectangle is 85 cm. express the perimeter of the rectangle is function of the width x.

Answers

the \ width : \ x \n the \ length : \ y \n The area : \ A=85 \n \nA=xy \n \n85=xy\n \ny= (85)/(x)


f(x)=85-2(y)  I think it is so .

What is the sum of the last four terms of the series 7+9+11+...+21

Answers

7 - 9 = 2 9 - 11 = 2 You should see that the series just increases by +2 therefore you can figure out what the last four terms are and then add them up one of the last terms given is 21

Solve to three significant digits 123= 500e ^ -0.12x

Answers

500{ e }^( -0.12x )=123\n \n { e }^{ -\frac { 12 }{ 100 } x }=\frac { 123 }{ 500 } \n \n { e }^{ -\frac { 3 }{ 25 } x }=\frac { 123 }{ 500 }

\n \n { \left( { e }^( x ) \right)  }^{ -\frac { 3 }{ 25 }  }=\frac { 123 }{ 500 } \n \n { e }^( x )={ \left( \frac { 123 }{ 500 }  \right)  }^{ \frac { 1 }{ -\frac { 3 }{ 25 }  }  }

\n \n { e }^( x )={ \left( \frac { 123 }{ 500 }  \right)  }^{ -\frac { 25 }{ 3 }  }\n \n \ln { \left( { \left( \frac { 123 }{ 500 }  \right)  }^{ -\frac { 25 }{ 3 }  } \right)  } =x

\n \n x=-\frac { 25 }{ 3 } \ln { \left( \frac { 123 }{ 500 }  \right)  }

\therefore \quad x\approx 11.7