All categories

3 years ago

Write the smallest possible odd number with all the following dight 5,2,3,7.

Mathematics

1 answer:

Lubov Fominskaja [6]3 years ago

5 0

Answer:

2357

Step-by-step explanation:

You might be interested in

One number is 4 more than another, and their sum is 60. What is the smaller number? If x = the larger number and y = the smaller

erik [133]

Answer:

a. $x+y=60$ and $-x+y=4$

Step-by-step explanation:

Let x be the larger number and y be the smaller number.

We have been given that one number is 4 more than another. We can represent this information in an equation as:

$y=x+4...(1)$

We can manipulate this equation as:

$y-x=x-x+4$

$-x+y=4$

We are also told that the sum of both numbers is 60. We can represent this information in an equation as:

$x+y=60...(2)$

Upon looking at our given choices, we can see that option 'a' is the correct choice.

8 0

3 years ago

Read 2 more answers

The rate of change of the downward velocity of a falling object is the acceleration of gravity (10 meters/sec 2) minus the accel

Alexus [3.1K]

Answer:

See below

Step-by-step explanation:

Write the initial value problem and the solution for the downward velocity for an object that is dropped (not thrown) from a great height.

if v(t) is the speed at time t after being dropped, v'(t) is the acceleration at time t, so the the initial value problem for the downward velocity is

v'(t) = 10 - 0.1v(t)

v(0) = 0 (since the object is dropped)

The equation v'(t)+0.1v(t)=10 is an ordinary first order differential equation with an integrating factor

$\bf e^{\int {0.1dt}}=e^{0.1t}$

so its general solution is

$\bf v(t)=Ce^{-0.1t}+100$

To find C, we use the initial value v(0)=0, so C=-100

and the solution of the initial value problem is

$\bf \boxed{v(t)=-100e^{-0.1t}+100}$

what is the terminal velocity?

The terminal velocity is

$\bf \lim_{t \to\infty}(-100e^{-0.1t}+100)=100\;mt/sec$

How long before the object reaches 90% of terminal velocity?

90% of terminal velocity = 90 m/sec

we look for a t such that

$\bf -100e^{-0.1t}+100=90\rightarrow -100e^{-0.1t}=-10\rightarrow e^{-0.1t}=0.1\\-0.1t=ln(0.1)\rightarrow t=\frac{ln(0.1)}{-0.1}=23.026\;sec$

How far has it fallen by that time?

The distance traveled after t seconds is given by

$\bf \int_{0}^{t}v(t)dt$

So, the distance traveled after 23.026 seconds is

$\bf \int_{0}^{23.026}(-100e^{-0.1t}+100)dt=-100\int_{0}^{23.026}e^{-0.1t}dt+100\int_{0}^{23.026}dt=\\-100(-e^{-0.1*23.026}/0.1+1/0.1)+100*23.026=1,402.6\;mt$

3 0

3 years ago

Gas Use

Alisiya [41]

Answer: 10.75
Steps:
After 110miles, it used 2.75gallons
Therefore, the amount of gas used per mile is: 100/2.75= 40 gallon per mile

So after 430miles, the number of gallons used: 430/40 = 10.75 gallons

Hope this helps ʕ•ᴥ•ʔ

7 0

3 years ago

What is the surface are of a rectangular prism of the length is 18cm the width 6cm and the height 3cm

Harlamova29_29 [7]

The surface area of rectangular prism is 360 square inches

Solution:

Given that rectangular prism of the length is 18 cm the width 6 cm and the height 3 cm

To find: Surface area of rectangular prism

The Surface area of rectangular prism is given by formula:

$A=2(wl+hl+hw)$

Where,

"l" is the length and "h" is the height and "w" is the width of prism

From given,

l = 18 cm

w = 6 cm

h = 3 cm

Substituting the values in formula,

$A = 2((6 \times 18) + (3 \times 18) + (3 \times 6))\\\\A = 2(108+54+18)\\\\A = 2 \times 180\\\\A = 360$

Thus surface area of rectangular prism is 360 square inches

4 0

3 years ago

Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2

Basile [38]

The Laplace transform of the given initial-value problem

$y' 5y = e^{4t}, y(0) = 2$ is mathematically given as

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is mathematically given as

$&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}$

$\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}$

$put $s=-s \Rightarrow a_{1}=\frac{17}{9}$$

$\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}$

In conclusion, Taking inverse Laplace tranoform

$L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\$

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

Write the smallest possible odd number with all the following dight 5,2,3,7.​

Write the smallest possible odd number with all the following dight 5,2,3,7.