Solve the following math problems: 6. Sketch graphs to show these inequalities: a) $x^2 \geq -4$ b) $x < 5$ 7. a) Solve the inequality $7 \geq 5x - 13$. b) Sketch a graph of the in... Solve the following math problems: 6. Sketch graphs to show these inequalities: a) $x^2 \geq -4$ b) $x < 5$ 7. a) Solve the inequality $7 \geq 5x - 13$. b) Sketch a graph of the inequality. 8. Given that x is an integer, solve the inequality: $x - 9 < 6x + 19$ 9. Solve these inequalities: a) $\frac{1}{-3}y \leq -5$ b) $8 - 3b > 41$ 10. The angles of a triangle are $x^\circ$, $x^\circ$ and $y^\circ$, where $y > 90^\circ$: a) What kind of triangle is it? b) Write down an inequality in terms of $x$. c) Write down an inequality in terms of $x$ and $y$. d) What is the range of values of $x$? Read more

Steps to Solve

1. Solve inequality 6a

$x^2 \geq -4$

Since $x^2$ is always non-negative for any real number $x$, $x^2$ will always be greater than or equal to $-4$. Thus, the solution is all real numbers.

2. Sketch the graph for inequality 6a

A number line with the entire line shaded to indicate all real numbers.

3. Solve inequality 6b

$x < 5$

4. Sketch the graph for inequality 6b

A number line with an open circle at $5$ and the line shaded to the left of $5$.

5. Solve the inequality 7a

$7 \geq 5x - 13$

Add $13$ to both sides:

$7 + 13 \geq 5x$

$20 \geq 5x$

Divide both sides by $5$:

$4 \geq x$ or $x \leq 4$

6. Sketch the graph for inequality 7b

A number line with a closed circle at $4$ (or a square bracket) and the line shaded to the left of $4$.

7. Solve the inequality 8

$x - 9 < 6x + 19$

Subtract $x$ from both sides:

$-9 < 5x + 19$

Subtract $19$ from both sides:

$-9 - 19 < 5x$

$-28 < 5x$

Divide both sides by $5$:

$-\frac{28}{5} < x$ or $x > -5.6$

Since $x$ is an integer, $x \geq -5$.

8. Solve the inequality 9a

$-\frac{1}{3y} \leq -5$

Multiply both sides by $-1$ (and reverse the inequality sign):

$\frac{1}{3y} \geq 5$

Since we don't know the sign of $y$, we will consider two different cases

Case 1: $y>0$, $3y>0$ Multiply both sides with $3y$: $1 \geq 15y$ $y \leq \frac{1}{15}$ Since we also have that $y>0$, that means $0 < y \leq \frac{1}{15}$

Case 2: $y<0$, $3y<0$ Multiply both sides with $3y$:

$1 \leq 15y$

$y \geq \frac{1}{15}$ The solution to this equation is invalid since we also have that $y<0$

Answer to $\frac{-1}{3y} \leq -5$: $0 < y \leq \frac{1}{15}$

9. Solve the inequality 9b

$8 - 3b > 41$

Subtract $8$ from both sides:

$-3b > 41 - 8$

$-3b > 33$

Divide both sides by $-3$ (and reverse the inequality sign):

$b < -11$

10. Analyze the triangle in 10

The angles of the triangle are $x, x, \text{ and } y^2$, where $y > 90^\circ$.

a. Since two angles are equal ($x$ and $x$), the triangle is an isosceles triangle.

b. The sum of angles in a triangle is $180^\circ$. Therefore, $x + x + y^2 = 180$, which simplifies to $2x + y^2 = 180$. Since $y > 90$ is not possible due to $2x + y^2 = 180$, we can assume that $x, x, \text{ and } y$ are the three angles where $y$ is in degrees and $y>0$. Using this assumption, we can state $x>0$.

c. Since the sum of angles is $180$: $2x + y^2 = 180$

d. Since $y > 0$, $y^2 > 0$. Also, each angle must be less than 180. $2x < 180$

$x < 90$

We know that $x$ and $y^2$ are angles of a triangle then they must both be greater than zero. Thus we can say $2x = 180 - y^2$

Thus, $x = 90 - \frac{y^2}{2}$

Since $y>0$, the largest value $y$ can take is when $x=0$, so $y = \sqrt{180}$. This is not possible, since $2x+y^2 = 180$ implies $x < 90$. Then $0 < x < 90$