Simplify this expressions 2sin30cos30

A jet travels 2301 miles against the wind in 3 hours and 2811 miles with the wind in the same amount of time. What is the rate o

astraxan [27]

Answer:

The rate of the jet in still air is 852 miles per hour. The rate of the wind is 85 miles per hour.

Step-by-step explanation:

Let suppose that jet travels uniformly, that is, at constant speed, the expressions for its travels against the wind and with the wind are, respectively:

Against the wind

$v -u = \frac{\Delta x_{1}}{\Delta t_{1}}$

With the wind

$v +u = \frac{\Delta x_{2}}{\Delta t_{2}}$

Where:

$v$ - Speed of the jet in still air, measured in miles per hour.

$u$ - Speed of wind, measured in miles per hour.

$\Delta x_{1}$ , $\Delta x_{2}$ - Distances travelled by jet against the wind and with the wind, measured in miles.

$\Delta t_{1}$ , $\Delta t_{2}$ - Times against the wind and with the wind, measured in hours.

By adding both expressions:

$2\cdot v = \frac{\Delta x_{1}}{\Delta t_{1}}+\frac{\Delta x_{2}}{\Delta t_{2}}$

$v = \frac{1}{2}\cdot \left(\frac{\Delta x_{1}}{\Delta t_{1}} + \frac{\Delta x_{2}}{\Delta t_{2}} \right)$

Given that $\Delta x_{1} = 2301\,mi$ , $\Delta t_{1} = 3\,h$ , $\Delta x_{2} = 2811\,mi$ and $\Delta t_{2} = 3\,h$ , the speed of the jet is:

$v = \frac{1}{2}\cdot \left(\frac{2301\,mi}{3\,h}+\frac{2811\,mi}{3\,h} \right)$

$v = 852\,\frac{mi}{h}$

The rate of the jet in still air is 852 miles per hour.

Lastly, the rate of the wind is:

$u = \frac{\Delta x_{2}}{\Delta t_{2}}-v$

$u = \frac{2811\,mi}{3\,h}-852\,\frac{mi}{h}$

$u = 85\,\frac{mi}{h}$

The rate of the wind is 85 miles per hour.

5 0

3 years ago

Suppose a reference variable of type Integer called myInt is already declared. Create an object of type Integer with the initial

Julli [10]

Answer:

//The class called "Solution" is declared

public class Solution {

//Main method which signify the beginning of program execution

public static void main(String[] args) {

//myInt variable of type Integer is declared

Integer myInt;

// An object of type Integer is declared with initial value of 1

Integer newInt = new Integer(1);

// The value of the new declared and assigned object is displayed to the user

System.out.println(newInt);

// The new declared object is assigned to the reference variable myInt

myInt = newInt;

// The value of myInt is displayed to the user

System.out.println(myInt);

}

Step-by-step explanation:

7 0

3 years ago

Answer the following ratios <br><br> 55 paise to Rs 1<br> 500ml to 2 liters

AfilCa [17]

55:1

500:2

There ya go!

7 0

3 years ago

How would you use the Fundamental Theorem of Calculus to determine the value(s) of b if the area under the graph g(x)=4x between

-BARSIC- [3]

The Fundamental Theorem of Calculus regarding geometry states that
$\int\limits^b_a g{(x)} \, dx = F(b)-F(a)$

Where F is the indefinite integral of $g(x)$

The first step is to integrate $g(x)$
$\int\ {4x} \, dx = \frac{4x^{1+1} }{1+1} = \frac{4x^{2} }{2} =2 x^{2}$

Then substitute the value of $b$ and $a=1$ into $2x^{2}$

$[2 (b)^{2}]-[2 (1)^{2}] = 240$
$2b^{2} -2=240$
$2b^{2}=240+2$
$2b^{2}=242$
$b^{2}= \frac{242}{2}$
$b^{2}=121$
$b=11$

Hence the limit of the area under $g(x)$ is between $a=1$ and $b=11$

6 0

3 years ago

Mr jones is losing 20 percent of his hair each year if he currently has 1546 hairs on about how many hair will he have left afte

dedylja [7]

I seemed to have forgot the formula for how to solve such a question so what I did was take my trust calculator, place the number he'd start with each year, multiply it by 0.2 or the 20%, and subtract that from the number he started with to get how many hairs he would start with in the next year. Then rinse-repeat until I got to 10 years.

Example:

1st year would be 1546 - (1546 x 0.2) = 1546 - 309.2 = 1236.8 hairs to start the 2nd year.

At the end of year 10, my calculation had Mr. Jones left with 166.000486 or ≈166 hairs left on his head. I'd suggest Mr. Jones invested in a women's razor; I've heard their blades are better for shaving yourself bald.

7 0

3 years ago

Simplify this expressions 2sin30cos30 ​

Simplify this expressions 2sin30cos30