Is 3.13 greater than 3.12

Mathematics

2 answers:

kkurt [141]3 years ago

7 0

Answer:

Yes

Step-by-step explanation:

13 is greater than 12 so

3.13>3.12

hoped this helped... :)

katen-ka-za [31]3 years ago

3 0

Answer:

yes

Step-by-step explanation:

if we figure out the difference between the two numbers by subtracting one from the other, if the product comes out positive, then the number you subtracted from is greater, if the product comes up negative, then the number you subtracted from is lesser

say we did 3.13 as our first number

3.13 - 3.12 = 0.01

this means that 3.13 is greater by 0.01, if we switch the two numbers however...

3.12 - 3.13 = -0.01

then that means that 3.12 is lesser by 0.01

i hope this makes sence, working with decimals may get tricky sometimes lol

You might be interested in

4 yards to 30 inches write each ratio in simplest form

MissTica

Answer:

24:5

Step-by-step explanation:

4 yards is 3 feet, so 4*3=12. There are 12 inches in a foot, so 12*12=144. 144:30= 72:15 = 24:5

6 0

2 years ago

Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

Answer:

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.