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babymother [125]
3 years ago
14

Select the false statement

Mathematics
2 answers:
Vlad1618 [11]3 years ago
5 0

Answer: D. Angles that form a linear pair are <u>sometimes</u> supplementary.

<em>That is not true, linear pairs are </em><em><u>always</u></em><em> supplementary.</em>

lbvjy [14]3 years ago
5 0

Answer:

D. Angles that form a linear pair are sometimes supplementary.

~OAmalOHopeO

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1. Where are you from

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3 years ago
Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1
Nesterboy [21]

Answer:

x=4

Step-by-step explanation:

Given \displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1, start by squaring both sides to work towards isolating x:

\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2

Recall (a-b)^2=a^2-2ab+b^2 and \sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}:

\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1

Isolate the radical:

\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}

Square both sides:

\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2

Expand using FOIL and (a+b)^2=a^2+2ab+b^2:

\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36

Move everything to one side to get a quadratic:

\displaystyle-\frac{1}{4}x^2+7x-24=0

Solving using the quadratic formula:

A quadratic in ax^2+bx+c has real solutions \displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}. In \displaystyle-\frac{1}{4}x^2+7x-24, assign values:

\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24

Solving yields:

\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}

Only x=4 works when plugged in the original equation. Therefore, x=24 is extraneous and the only solution is \boxed{x=4}

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