List the following numbers from least to greatest<br><br> -5, 0, 2, -20 -6,1,0,-15

GarryVolchara [31]

See the answer in the image

4 0

2 years ago

The volume V (in cubic feet) of a right cylinder with a height of 6 feet and radius r (in feet) is given by V=6πr2 . Solve the

Zinaida [17]

Answer:

a) r = √V/6π

b) r = 8 inches

Step-by-step explanation:

a) The volume V (in cubic feet) of a right cylinder with a height of 6 feet and radius r (in feet) is given by V=6πr2 . Solve the formula for r .

Solving for r means making r the subject of the formula

= V=6πr²

Divide both side by 6π

V/6π = 6πr²/6π

r² = V/6π

r = √V/6π

b) Then find the radius of the cylinder when the volume is 1206 cubic feet. Round your answer to the whole number.

r = √V/6π

V = 1206 ft³

r = √1206/6π

r = √(63.98028712294193)

r = 7.9987678503 inches

Approximately r = 8 inches

5 0

3 years ago

Given the second order homogeneous constant coefficient equation y′′−4y′−12y=0 1) the characteristic polynomial ar2+br+c is r^2-

Vitek1552 [10]

Answer:

The initial value problem $y(x) = 4 e^{-2x} -3 e^{6 x}$

Step-by-step explanation:

<u>Step1:</u>-

a) Given second order homogenous constant co-efficient equation

$y^{ll} - 4y^{l}-12y=0$

Given equation in the operator form is $(D^{2} -4D-12)y=0$

<u>Step 2</u>:-

b) Let f(D) = $(D^{2} -4D-12)y$

Then the auxiliary equation is $(m^{2} -4m-12)=0$

Find the factors of the auxiliary equation is

$m^{2} -6m+2m-12=0$

m(m-6) + 2(m-6) =0

m+2 =0 and m-6=0

m=-2 and m=6

The roots are real and different

The general solution $y = c_{1} e^{-m_{1} x} + c_{2} e^{m_{2} x}$

the roots are $m_{1} = -2 and m_{2} = 6$

The general solution of given differential equation is

$y = c_{1} e^{-2x} + c_{2} e^{6 x}$

<u>Step 3</u>:-

C) Given initial conditions are y(0) =1 and y1 (0) =-26

The general solution of given differential equation is

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$ .....(1)

substitute x =0 and y(0) =1

$y(0) = c_{1} e^{0} + c_{2} e^{0}$

$1 = c_{1} + c_{2}$ .........(2)

Differentiating equation (1) with respective to 'x'

$y^l(x) = -2c_{1} e^{-2x} + 6c_{2} e^{6 x}$

substitute x= o and y1 (0) =-26

$-26 = -2c_{1} e^{0} + 6c_{2} e^{0}$

$-2c_{1} + 6c_{2} = -26$ .............(3)

solving (2) and (3) by using substitution method

$substitute c_{2} =1- c_{1}$ in equation (3)

$-2c_{1} + 6(1-c_{1}) = -26$

on simplification , we get

$-2c_{1} + 6(1)-6c_{1}) = -26$

$-8c_{1} = -32$

dividing by'8' we get $c_{1} =4$

substitute $c_{1} =4$ in equation $1 = c_{1} + c_{2}$

so $c_{2} = 1-4 = -3$

now substitute $c_{1} =4 and c_{2} =-3$ in general solution

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$

$y(x) = 4 e^{-2x} -3 e^{6 x}$

now the initial value problem

$y(x) = 4 e^{-2x} -3 e^{6 x}$

7 0

4 years ago

Which of the following is a correct interpretation of the expression -4 + 13−4+13minus, 4, plus, 13?

MaRussiya [10]

the sum between negative 4 and 13, this can be tought as:

Stepping on -4 and moving 13 units to the right.
Stepping on 13 and moving 4 units to the left.

<h3>Which is the interpretation of the given expression?</h3>

The expression is just the sum:

-4 + 13

So we have the sum between negative 4 and 13, this can be tought as:

Stepping on -4 and moving 13 units to the right.
Stepping on 13 and moving 4 units to the left.

Now, none of the written options (a, b, and c) depict any of this, so I assume that the correct option is the one that you did not write, which is option D.

If you want to learn more about expressions:

brainly.com/question/4344214

#SPJ1

3 0

1 year ago

Which of the following are solutions to the system

Tomtit [17]

Answer:

A & B

Step-by-step explanation:

5x - 2y = 10 (1)

-2.5x + y = -5 (2)

When you multiply (2) by -2, you get (1), so there can be many solutions

To choose from the given options, plug in the values to check which ones satisfy the equation

5x - 2y = 10

A) (2,0)

5(2) - 2(0) = 10

10 = 10 (true statement)

B) (-2,-10)

5(-2) - 2(-10) = 10

-10 + 20 = 10

10 = 10 (true statement)

5 0

3 years ago

List the following numbers from least to greatest -5, 0, 2, -20 -6,1,0,-15