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3 years ago

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Elizabeth met 24 of her cousins at a family reunion. the number of male cousins m was 6 less than twice the number of female cou

sins f . write a system of equations to find the number of male cousins and female cousins

1 answer:

atroni [7]3 years ago

5 0

There are 14 male cousins and 10 female cousins.

Step-by-step explanation:

Let,

m be the male cousins and f be the female cousins

According to given statement;

m+f =24 Eqn 1

m= 2f - 6 Eqn 2

Putting value of m in Eqn 1

$2f-6+f=24\\3f=24+6\\3f=30$

Dividing both sides by 3

$\frac{3f}{3}=\frac{30}{3}\\f=10$

Putting f=10 in Eqn 1

$m+10=24\\m=24-10\\m=14$

There are 14 male cousins and 10 female cousins.

Keywords: Linear equations, substitution method

Learn more about linear equations at:

#LearnwithBrainly

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Please help!! What is the solution to the quadratic inequality? 6x2≥10+11x

fredd [130]

Answer:

The solution of the inequation $6\cdot x^{2} \geq 10 + 11\cdot x$ is $\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)$ .

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

$6\cdot x^{2} \geq 10 + 11\cdot x$

$6\cdot x^{2}-11\cdot x -10 \geq 0$

$6\cdot \left(x^{2}-\frac{11}{6}\cdot x -\frac{10}{6} \right)\geq 0$

Roots are found by Quadratic Formula:

$r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}$

$r_{1} = \frac{5}{2}$ and $r_{2} = -\frac{2}{3}$

Then, the factorized form of the inequation is:

$6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0$

By Real Algebra, there are two condition that fulfill the inequation:

a) $x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0$

$x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}$

$x \geq \frac{5}{2}$

b) $x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0$

$x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}$

$x\leq -\frac{2}{3}$

The solution of the inequation $6\cdot x^{2} \geq 10 + 11\cdot x$ is $\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)$ .

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$\huge \boxed{\mathbb{QUESTION} \downarrow}$

What is the solution to the inequality -3x-42>3 ?

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

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