A package wieghs 96 ounces. what is the weight of the pa kage in pounds?

Could anyone help me with this? (A question I've been wondering about.)

Kamila [148]

Xbxhfjtiifixkdjdyfhdkkd 388484

5 0

3 years ago

The rectangle shown has a perimeter of 146 cm and the given area. Its length is 7 more than five times its width. Write and solv

dangina [55]

Answer:

System of equations:

L = 5W + 7

2W + 2L = P

L = 62 cm

W = 11 cm

Step-by-step explanation:

Given the measurements and key words/phrases in the problem, we can set up two different equations that can be used to find both variables, length and width, of the rectangle.

The formula for perimeter of a rectangle is: 2W + 2L = P, where W = width and L = length. We also know that the L is '7 more than five times its width'. This can be written as: L = 5W + 7. Using this expression for the value of 'L', we can use the formula for perimeter and solve for width:

2W + 2(5W + 7) = 146

Distribute: 2W + 10W + 14 = 146

Combine like terms: 12W + 14 = 146

Subtract 14 from both sides: 12W + 14 - 14 = 146 - 14 or 12W = 132

Divide 12 by both sides: 12W/12 = 132/12 or W = 11

Put '11' in for W in the equation for 'L': L = 5(11) + 7 or L = 55 + 7 = 62.

5 0

3 years ago

A high school drama club is selling tickets for a fundraiser event. Based on data from past events, the number of tickets sold c

-Dominant- [34]

Answer:

600 or 16

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A robot can complete 6 tasks in 7/10 hour. Each task takes the same amount of time. a. How long does it take the robot to comple

Jlenok [28]

Answer:

7/60 hour or 7 minutes b) 8 or 8 4/7 depending if you consider a fraction of a task is yes or no

Step-by-step explanation:

6 0

2 years ago

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

=0

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

=0

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

7 0

2 years ago