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3 years ago

9

At the sports awards ceremony,5 members of the soccer team and 8 members of the softball team won awards. What was the ratio of

the softball team members to award winners?

1 answer:

Dima020 [189]3 years ago

8 0

Answer:

8:13 or 8/13 or 62%

Step-by-step explanation:

When comparing two things make sure to put what the question says first, first. Add the number of soccer and softball players that got awards to find the total number of award winners.

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3 years ago

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$

We use the initial condition $y(0)=-12$ to find the value of C.

$-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}$

Therefore,

$y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}$

4 0

3 years ago

LCM of x<sup>2</sup>+5x+6 and x<sup>2</sup>-x-6 is ………………………

sergejj [24]

Answer:

$(x^2 - 9)(x + 2)$

Step-by-step explanation:

Given:

$x^2 + 5x + 6$

$x^2 - x - 6$

Required:

LCM of the polynomials

SOLUTION:

Step 1: Factorise each polynomial

$x^2 + 5x + 6$

$x^2 + 3x + 2x + 6$

$(x^2 + 3x) + (2x + 6)$

$x(x + 3) + 2(x + 3)$

$(x + 2)(x + 3)$

$x^2 - x - 6$

$x^2 - 3x +2x - 6$

$x(x - 3) + 2(x - 3)$

$(x + 2)(x - 3)$

Step 2: find the product of each factor that is common in both polynomials.

We have the following,

$x^2 + 5x + 6 = (x + 2)(x + 3)$

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

The common factors would be: =>

$(x + 2)$ (this is common in both polynomials, so we would take just one of them as a factor.

$(x + 3)$ and,

$(x - 3)$

Their product = $(x - 3)(x + 3)(x +2) = (x^2 - 9)(x + 2)$

4 0

3 years ago