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3 years ago

Which number line represents the solution set for the inequality 3x<-9

Mathematics

2 answers:

KiRa [710]3 years ago

6 0

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$3x < -9$

solve for x

Divide by $3$ both sides

$3x/3 < -9/3$

$x < -3$

The solution is the interval------> $(-infinite, -3)$ -

All real numbers less than $-3$

In a number line the solution is the shaded area to the left of the dashed line (open circle)

see the attached figure

r-ruslan [8.4K]3 years ago

3 0

The answer would be the 3rd number line or "C"

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Complete the identity. 1) sec^4 x + sec^2 x tan^2 x - 2 tan^4 x = ?

Alecsey [184]

Answer:

See Explanation

Step-by-step explanation:

Question like this are better answered if there are list of options; However, I'll simplify as far as the expression can be simplified

Given

$sec^4 x + sec^2 x tan^2 x - 2 tan^4 x$

Required

Simplify

$(sec^2 x)^2 + sec^2 x tan^2 x - 2 (tan^2 x)^2$

Represent $sec^2x$ with a

Represent $tan^2x$ with b

The expression becomes

$a^2 + ab- 2 b^2$

Factorize

$a^2 + 2ab -ab- 2 b^2$

$a(a + 2b) -b(a+ 2 b)$

$(a -b) (a+ 2 b)$

Recall that

$a = sec^2x$

$b = tan^2x$

The expression $(a -b) (a+ 2 b)$ becomes

$(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

..............................................................................................................................

In trigonometry

$sec^2x =1 +tan^2x$

Subtract $tan^2x$ from both sides

$sec^2x - tan^2x =1 +tan^2x - tan^2x$

$sec^2x - tan^2x =1$

..............................................................................................................................

Substitute 1 for $sec^2x - tan^2x$ in $(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

$(1) (sec^2x+ 2 tan^2x)$

Open Bracket

$sec^2x+ 2 tan^2x$ ------------------This is an equivalence

$(secx)^2+ 2 (tanx)^2$

Solving further;

................................................................................................................................

In trigonometry

$secx = \frac{1}{cosx}$

$tanx = \frac{sinx}{cosx}$

Substitute the expressions for secx and tanx

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$(secx)^2+ 2 (tanx)^2$ becomes

$(\frac{1}{cosx})^2+ 2 (\frac{sinx}{cosx})^2$

Open bracket

$\frac{1}{cos^2x}+ 2 (\frac{sin^2x}{cos^2x})$

$\frac{1}{cos^2x}+ \frac{2sin^2x}{cos^2x}$

Add Fraction

$\frac{1 + 2sin^2x}{cos^2x}$ ------------------------ This is another equivalence

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In trigonometry

$sin^2x + cos^2x= 1$

Make $sin^2x$ the subject of formula

$sin^2x= 1 - cos^2x$

................................................................................................................................

Substitute the expressions for $1 - cos^2x$ for $sin^2x$

$\frac{1 + 2(1 - cos^2x)}{cos^2x}$

Open bracket

$\frac{1 + 2 - 2cos^2x}{cos^2x}$

$\frac{3 - 2cos^2x}{cos^2x}$ ---------------------- This is another equivalence

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3 years ago

Mikey bought 4 of the 500 raffle tickets sold to

Zina [86]

Answer:

4/500

Step-by-step explanation:

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3 years ago

2 If A= 4 3 find A1 using elementary row operations.

MA_775_DIABLO [31]

Answer: $A^{-1}=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]$

Step-by-step explanation:

$\left[\begin{array}{cc}2&1\\4&3\end{array}\right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

$\dfrac{1}{2}Row\ 1\rightarrow\left[\begin{array}{cc}1&\frac{1}{2}\\4&3\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\0&1\end{array}\right]$

$Row\ 2 -4 \ Row\ 1\rightarrow \left[\begin{array}{cc}1&\frac{1}{2}\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\-2&1\end{array}\right]$

$Row\ 1-\dfrac{1}{2}\ Row\ 2 \rightarrow \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]$

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