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3 years ago

2/3 greater than or less than 1/5

2 answers:

5 0

Greater. Imagine it as a pie you would rather 2 big slices of pie on a 3 slice pie, than 1 skinny slice on a 5 piece pie.

UkoKoshka [18]3 years ago

4 0

2/3 is greater. 1/5 is less.

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Add<br> 6 3/5 + (-6 1/3<br> Enter your answer as a simplified fraction in the box

Sholpan [36]

Okay so basically the answer would be 4/15

8 0

2 years ago

Flora has 33 flowers and she can pick 7 flowers a minute. Write an equation that shows how many flowers, F, that Flora will have

monitta

F=7M
33=M
Both answers will work

3 0

3 years ago

louise bought 80 metres of fencing to make an enclosure for her pet dog .tommy if louise expects a rectangular enclosure what is

OleMash [197]

Answer:

400 m^2.

Step-by-step explanation:

The largest area is obtained where the enclosure is a square.

I think that's the right answer because a square is a special form of a rectangle.

So the square would be 20 * 20 = 400 m^2.

Proof:

Let the sides of the rectangle be x and y m long

The area A = xy.

Also the perimeter 2x + 2y = 80

x + y = 40

y = 40 - x.

So substituting for y in A = xy:-

A = x(40 - x)

A = 40x - x^2

For maximum value of A we find the derivative and equate it to 0:

derivative A' = 40 - 2x = 0

2x = 40

x = 20.

So y = 40 - x

= 40 - 20

=20

x and y are the same value so x = y.

Therefore for maximum area the rectangle is a square.

3 0

3 years ago

What’s the answer pls

MAXImum [283]

37.1
35 * .06 = 2.1
35 + 2.1 = 37.1

8 0

3 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

3 years ago

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