Question

Please answer all three

Answer 1

                       Question # 7

Answer:

We conclude that the statement B is true. The solution is also attached below.

Step-by-step Explanation:

As the inequality graphed on the number line showing that solution must be   < x    (-∞, 3] U [5, ∞)

So, lets check the statements to know which statement has this solution.

Analyzing statement A)

$x^2-3x+5>\:0$

$\mathrm{Write}\:x^2-3x+5\:\mathrm{in\:the\:form:\:\:}x^2+2ax+a^2$

$2a=-3\quad :\quad a=-\frac{3}{2}$

$\mathrm{Add\:and\:subtract}\:\left(-\frac{3}{2}\right)^2\:$

$x^2-3x+5+\left(-\frac{3}{2}\right)^2-\left(-\frac{3}{2}\right)^2$

$\mathrm{Complete\:the\:square}$

$\left(x-\frac{3}{2}\right)^2+5-\left(-\frac{3}{2}\right)^2$

$\mathrm{Simplify}$

$\left(x-\frac{3}{2}\right)^2+\frac{11}{4}$

So,

$\left(x-\frac{3}{2}\right)^2>-\frac{11}{4}$

Thus,

$x^2-3x+5>0\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:\mathrm{True\:for\:all}\:x\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}$

Therefore, option A) is FALSE.

Analyzing statement B)

(x + 3) (x - 5) ≥ 0

$x^2-2x-15\ge 0$

$\left(x+3\right)\left(x-5\right)=0$       $\left(Factor\:left\:side\:of\:equation\right)$

$x+3=0\:or\:x-5=0$

$x=-3\:or\:x=5$

So

$x\le \:-3\quad \mathrm{or}\quad \:x\ge \:5$

Thus,

$\left(x+3\right)\left(x-5\right)\ge \:0\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-3\quad \mathrm{or}\quad \:x\ge \:5\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-3]\cup \:[5,\:\infty \:)\end{bmatrix}$

Therefore, the statement B is true.

Solution is also attached below.

Analyzing statement C)

$x^2+2x-15\ge 0$

$\mathrm{Factor}\:x^2+2x-15:\quad \left(x-3\right)\left(x+5\right)$

So,

$x\le \:-5\quad \mathrm{or}\quad \:x\ge \:3$

$x^2+2x-15\ge \:0\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-5\quad \mathrm{or}\quad \:x\ge \:3\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-5]\cup \:[3,\:\infty \:)\end{bmatrix}$

Therefore, option C) is FALSE.

Analyzing statement D)

- 3 < x < 5

$-3$

Therefore, option D) is FALSE.

Analyzing statement E)

None of the above

The statement E) is False also as the statement B represents the correct solution.

Therefore, from the discussion above, we conclude that the statement B is true. The solution is also attached below.

                         Question # 8

Find the number that is $\frac{1}{3}$ of the way from $\:2\frac{1}{6}$ to $\:5\frac{1}{4}$.

Answer:

Therefore, $\frac{37}{36}$ is the number that is  $\frac{1}{3}$ of the way from  $\:2\frac{1}{6}$ to $\:5\frac{1}{4}$.

Step-by-step Explanation:

$\mathrm{Convert\:mixed\:numbers\:to\:improper\:fraction:}\:a\frac{b}{c}=\frac{a\cdot \:c+b}{c}$

So,

$2\frac{1}{6}=\frac{13}{6}$

$5\frac{1}{4}=\frac{21}{4}$

As the length from $\frac{21}{4}$ to $\frac{13}{6}$ is

$\frac{21}{4}-\frac{13}{6}=\frac{37}{12}$

Now Divide $\frac{37}{12}$ into 3 equal parts. So,

$\frac{37}{12}\div \:3=\frac{37}{36}$

As we have to find number that is $\frac{1}{3}$ of the way from  $\:2\frac{1}{6}$ to $\:5\frac{1}{4}$, it means it must have covered 2/3 of the way. As we have divided  $\frac{37}{12}$ into 3 equal parts, which is $\frac{37}{36}$

Therefore, $\frac{37}{36}$ is the number that is  $\frac{1}{3}$ of the way from  $\:2\frac{1}{6}$ to $\:5\frac{1}{4}$.

                       Question # 9

Answer:

$\left(2x+3\right)$ is in the form $dx+\:e$.

Step-by-step Explanation:

Considering the expression

$2x^2+11x+12$

Factor

$2x^2+11x+12$

$\mathrm{Break\:the\:expression\:into\:groups}$

$\left(2x^2+3x\right)+\left(8x+12\right)$

$\mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2+3x\mathrm{:\quad }x\left(2x+3\right)$

$\mathrm{Factor\:out\:}4\mathrm{\:from\:}8x+12\mathrm{:\quad }4\left(2x+3\right)$

$x\left(2x+3\right)+4\left(2x+3\right)$

$\mathrm{Factor\:out\:common\:term\:}2x+3$

$\left(2x+3\right)\left(x+4\right)$

Therefore, $\left(2x+3\right)$ is in the form $dx+\:e$.

Keywords: factor, ratio, solution

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