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2 years ago

Jet travelling at 500 km/h for 4 hours.

Mathematics

2 answers:

stellarik [79]2 years ago

8 0

Answer: jet travels a total of 2000 km in total

Step-by-step explanation:

Sauron [17]2 years ago

7 0

Whats the question?

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the combined age of april and laura is 23 years. laura's age is two years more than half of april's age. what is laura's age

Kipish [7]

Laura's age is 9 years.

Solution:

Let x be the age of April.

Laura's age = 2 years more than half of April's age

<u>Convert statement into algebraic expression:</u>

Half of April's age = $\frac{x}{2}$

2 years more than half of April's age = $\frac{x}{2}+2$

Combined age of April and Laura = 23

⇒ April's age + Laura's age = 23

$$\Rightarrow \ \ x+(\frac{x}{2}+2) =23$

$$\Rightarrow \ \ x+\frac{x}{2}+2 =23$

$$\Rightarrow \ \ \frac{x}{1}+\frac{x}{2}+\frac{2}{1} =23$

To add the fractions make the denominators same.

Multiplying 2 on both numerator and denominator of unlike terms, we get

$$\Rightarrow \ \ \frac{x\times2}{1\times2}+\frac{x}{2}+\frac{2\times2}{1\times2} =23$

$$\Rightarrow \ \ \frac{2x}{2}+\frac{x}{2}+\frac{4}{2} =23$

Denominators are same, now add the fractions.

$$\Rightarrow \ \ \frac{2x+x+4}{2} =23$

Do cross multiplication.

$$\Rightarrow \ \ 2x+x+4=23\times2$

$$\Rightarrow \ \ 3x=46-4$

$$\Rightarrow \ \ 3x=42$

$$\Rightarrow \ \ x=14$

Aprils's age = 14 years

Laura's age = $\frac{14}{2}+2$

= 7 + 2

Laura's age = 9

Hence Laura's age is 9 years.

8 0

3 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago

A survey found that 78% of students do their homework before 10:00 predict how many students out 975 do their homework before 10

GenaCL600 [577]

Hi Zahra, I've found a very simple and effective way of calculating percentages. 78% refers to 78 out of every 100, which can also be written as "0.78". Now, we can simply multiply 0.78 by your number (975) to get 760.5 students. Obviously, there isn't a half a student, so we round that number up to 761. So 761 out of 975 students do their homework before 10:00.

4 0

3 years ago

Read 2 more answers

Im a little stuck on this (please show how you did it)

nikklg [1K]

Answer:

all work is shown / pictured