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3 years ago

I will name you the brainliest ! Which point represents -(-10) on the number line?

Mathematics

2 answers:

earnstyle [38]3 years ago

8 0

Answer:

Step-by-step explanation:

When you subtract a negative, it is the same thing as adding a positive
So, -(-10) = 10
E = 10, so E is the correct answer

I hope this helps!

Anna007 [38]3 years ago

4 0

Answer:

Step-by-step explanation:

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5x – 6 < 9 :P porfavore

Nikitich [7]

Answer:

$x$

Step-by-step explanation:

Treat this the same way you would an equation. Isolate the variable. Add 6 to both sides:

$5x-6+6$

Divide both sides by 5:

$\frac{5x}{5}$

The value of x is less than 3.

:Done

3 0

3 years ago

Read 2 more answers

Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$

Accordingly:

$x=-4 \ is \ a \ vertical \ asymptote \ for \\ \\ f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20}$

<h2>Learn more:</h2>

Vertical and horizontal asymptotes: brainly.com/question/10254973

#LearnWithBrainly

5 0

4 years ago

find the orthogonal projection of v= [19,12,14,-17] onto the subspace W spanned by [ [ -4,-1,-1,3] ,[ 1,-4,4,3] ] proj w (v) = [

12345 [234]

<h2>Answer:</h2>

Hence, we have:

$proj_W(v)=[\dfrac{464}{21},\dfrac{167}{21},\dfrac{71}{21},\dfrac{-131}{7}]$

<h2>Step-by-step explanation:</h2>

By the orthogonal decomposition theorem we have:

The orthogonal projection of a vector v onto the subspace W=span{w,w'} is given by:

$proj_W(v)=(\dfrac{v\cdot w}{w\cdot w})w+(\dfrac{v\cdot w'}{w'\cdot w'})w'$

Here we have:

$v=[19,12,14,-17]\\\\w=[-4,-1,-1,3]\\\\w'=[1,-4,4,3]$

Now,

$v\cdot w=[19,12,14,-17]\cdot [-4,-1,-1,3]\\\\i.e.\\\\v\cdot w=19\times -4+12\times -1+14\times -1+-17\times 3\\\\i.e.\\\\v\cdot w=-76-12-14-51=-153$

$w\cdot w=[-4,-1,-1,3]\cdot [-4,-1,-1,3]\\\\i.e.\\\\w\cdot w=(-4)^2+(-1)^2+(-1)^2+3^2\\\\i.e.\\\\w\cdot w=16+1+1+9\\\\i.e.\\\\w\cdot w=27$

and

$v\cdot w'=[19,12,14,-17]\cdot [1,-4,4,3]\\\\i.e.\\\\v\cdot w'=19\times 1+12\times (-4)+14\times 4+(-17)\times 3\\\\i.e.\\\\v\cdot w'=19-48+56-51\\\\i.e.\\\\v\cdot w'=-24$

$w'\cdot w'=[1,-4,4,3]\cdot [1,-4,4,3]\\\\i.e.\\\\w'\cdot w'=(1)^2+(-4)^2+(4)^2+(3)^2\\\\i.e.\\\\w'\cdot w'=1+16+16+9\\\\i.e.\\\\w'\cdot w'=42$

Hence, we have:

$proj_W(v)=(\dfrac{-153}{27})[-4,-1,-1,3]+(\dfrac{-24}{42})[1,-4,4,3]\\\\i.e.\\\\proj_W(v)=\dfrac{-17}{3}[-4,-1,-1,3]+(\dfrac{-4}{7})[1,-4,4,3]\\\\i.e.\\\\proj_W(v)=[\dfrac{68}{3},\dfrac{17}{3},\dfrac{17}{3},-17]+[\dfrac{-4}{7},\dfrac{16}{7},\dfrac{-16}{7},\dfrac{-12}{7}]\\\\i.e.\\\\proj_W(v)=[\dfrac{464}{21},\dfrac{167}{21},\dfrac{71}{21},\dfrac{-131}{7}]$