Which is bigger 2m or 275cm

Mathematics

2 answers:

aliina [53]1 year ago

8 0

Answer:

275.

Step-by-step explanation:

If you convert 2m into CM you get 200.
If you convert 275 cm into M you get 2.75. So 275cm is bigger.

Advocard [28]1 year ago

6 0

275 centimeters is bigger than 2 meters

You might be interested in

Can someone help me answer these questions

Vlad1618 [11]

We can't see the questions from this picture of you show a picture that we can see or type the answers we can answer

4 0

2 years ago

Shane's money is 3:11 if Isabella has 33 dollars how much do shane and Isabella have together

grin007 [14]

3:11
S:I
?:33
So if Isabella has 33 dollars that means they did 11x3=33 so to find Shanes you would do 3x3=9 then add the two together which is $42 hope this helps

3 0

3 years ago

Simplify (2√5+3√7)squared

belka [17]

Ok....
Simplified would be 12.40
Squared would be 153.99 or fraction form would be 83 + 12 $\sqrt{35}$

H0P3 It H3LPS :)

6 0

3 years ago

Which of the following statements does NOT adequately describe a developmental progression?

Artyom0805 [142]

The statement which does not adequately describe a developmental progression is Choice C- A developmental progression is a skill check- list that four-years- olds must master to be prepared for kindergarten.

<h3>What is developmental progression?</h3>

The Developmental Progression of Functional Skills is a concept which describes the child's growth in which case, the child's develops relationship with family members and a sense of self in the process and this concept in discuss is not limited to four-year olds.

Read more on developmental progression;

brainly.com/question/13670359

#SPJ1

6 0

1 year ago

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$