A chess club has 24 members, of which 1 is a male and the rest are females.

The sum of three certain consecutive numbers is 240. Find the three numbers. [Hint: Let x stand for the smallest number. Since t

aleksley [76]

Write an equation based on the problem in the terms of x and solve the equation
first number + second number + third number = 240
x + x + 1 + x + 2 = 240
3x + 3 = 240
3x = 240 - 3
3x = 237
x = 237/3
x = 79
The smallest number, x, is equal to 79

Find the next two numbers
the second number = x + 1
the second number = 79 + 1
the second number = 80

the third number = x + 2
the third number = 79 + 2
the third number = 81

ANSWER: The three numbers are 79, 80, 81

7 0

4 years ago

Which statement correctly compares the two functions? A. They have the same end behavior as x approaches [infinity], but they ha

OLga [1]

Answer:

A. They have the same end behavior as x approaches [Infinity], but they have different x- and y- intercepts.

Step-by-step explanation:

The x and y intercepts in mathematics can depict more than one function. They can have same end behavior as x approach. The x and y intercepts can be different for the two functions.

5 0

3 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.