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Rashid [163]
3 years ago
11

Solve the inequality for 1>x+6 .

Mathematics
1 answer:
Komok [63]3 years ago
7 0

Answer:

-5 > x

Explanation:

To do this, you have to use the inverse operations. Subtract 6 from each side. That equals -5. Bring everything down and that is your answer!

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Read 2 more answers
Cot^2x/cscx-1=1+sinx/sinx
KATRIN_1 [288]
\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-------------------------------\\\\
\cfrac{cot^2(x)}{csc(x)-1}=\cfrac{1+sin(x)}{sin(x)}\impliedby \textit{let's do the left-hand-side}

\bf \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1}{sin(x)}-1}\implies \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1-sin(x)}{sin(x)}}\implies \cfrac{cos^2(x)}{sin^2(x)}\cdot \cfrac{sin(x)}{1-sin(x)}
\\\\\\
\cfrac{cos^2(x)}{sin(x)}\cdot \cfrac{1}{1-sin(x)}\implies \cfrac{cos^2(x)}{sin(x)[1-sin(x)]}

\bf \cfrac{1-sin^2(x)}{sin(x)[1-sin(x)]}\implies \cfrac{1^2-sin^2(x)}{sin(x)[1-sin(x)]}
\\\\\\
\cfrac{\underline{[1-sin(x)]}~[1+sin(x)]}{sin(x)\underline{[1-sin(x)]}}\implies \cfrac{1+sin(x)}{sin(x)}
5 0
3 years ago
Which expression is equivalent to the expression below?
natulia [17]

Answer:

Step-by-step explanation:

3(7-22x) = 21-66x

3 0
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