All categories

4 years ago

Is 10/9 closer to 1,1/2,0

2 answers:

5 0

$\dfrac{10}{9}=\dfrac{9+1}{9}=\dfrac{9}{9}+\dfrac{1}{9}=1\dfrac{1}{9} \ \textgreater \ 1\\\\therefore\\\\\dfrac{10}{9}\ is\ closter\ to\ 1$

Doss [256]4 years ago

4 0

10/9 would be closer to 1 because 9/9 would be 1 whole and 10/9 is a little bigger than a whole
So it would be closer to 1

Hope this helped!! :D

You might be interested in

What is the surface area of this design? 5 in 5 in 6.4 in 5 in 9 in

xeze [42]

ANSWER

$197 {in}^{2}$

EXPLANATION

The area of the two tra-pezoidal faces

$= 2 \times \frac{1}{2} \times (9 + 5) \times 5 = 70 {in}^{2}$

The area of the 5 by 6.4 rectangular face

$= 6.4 \times 5 = 32 {in}^{2}$

The area of the two square faces

$= 2(5 \times 5) = 50 {in}^{2}$

The area of the 9 by 5 rectangular face is

$= 9 \times 5 = 45 {in}^{2}$

The surface area of the design is:

$70 + 32 + 50 + 45=197 {in}^{2}$

6 0

3 years ago

Help plzzzzzz 1. Use your expression from Exercise 15 to find the total cost of 8 play tickets and 8 choir concert tickets

Radda [10]

Answer:637

Step-by-step explanation:

7362 is the andwrr

8 0

3 years ago

Simplify: − 2 3 (−3 − 9x) A) −2 − 9x B) 2 − 9x C) 2 − 6x D) 2 + 6x

Stolb23 [73]

Answer:

2 +6x

Step-by-step explanation:

-2/3 (-3- 9x)

Distribute

-2/3*-3 -2/3 *-9x

2 +6x

4 0

3 years ago

An online book store has 182 boxes of books in its warehouse. Each box has 32 books. Part A Which equation represents the best e

kari74 [83]

Answer:

Do it yourself bro

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is

tresset_1 [31]

Answer:

a) $x(t) = 13*e^(^-^\frac{t}{100}^)$

b) 10.643 kg

Step-by-step explanation:

Solution:-

- We will first denote the amount of salt in the solution as x ( t ) at any time t.

- We are given that the Pure water enters the tank ( contains zero salt ).

- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min

- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.

- The ODE is mathematically expressed as:

$\frac{dx}{dt} =$ ( salt flow in ) - ( salt flow out )

- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.

- So any time ( t ) the concentration of salt in the 5,000 L is:

$conc = \frac{x(t)}{1000}\frac{kg}{L}$

- The amount of salt leaving the tank per unit time can be determined from:

salt flow-out = conc * V( flow-out )

salt flow-out = $\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\$

salt flow-out = $\frac{x(t)}{100}\frac{kg}{min}$

- The ODE becomes:

$\frac{dx}{dt} = 0 - \frac{x}{100}$

- Separate the variables and integrate both sides:

$\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)$

- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:

$13 = C*e^0 = C$

- The solution to the ODE becomes:

$x(t) = 13*e^(^-^\frac{t}{100}^)$

- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:

$x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg$

- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg

7 0

3 years ago