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4 years ago

Which inequality matches the graph?

Mathematics

2 answers:

inn [45]4 years ago

8 0

Answer:

D) x ≤ -2

Step-by-step explanation:

The dot is at -2 = -2

the dot is filled in = ≤ or ≥ because its means equal too

think of ≤ or ≥ like arrows = it points left.

so its x ≤ -2

Note on using the ≤ or ≥ like pointing arrows

this only works if its written:

variable < constant form,

which is the way they are usually seen, like x < 5

but it could be 5 > x which means the > points the opposite way

nata0808 [166]4 years ago

3 0

Hello! :)

Answer:

D. x≤-2

*The answer must have a negative sign and less than or equal to.*

Step-by-step explanation:

Lesson: It's about the graphing inequalities

An inequality is a statement that compares mathematical expressions that are not equal. The expressions are separated by an inequality symbol (<,>, ≤, ≥, or ≠).

The arrow points to the left. The circle is closed.

Hope this helps!

Have a nice day! :)

-Charlie

Thanks!

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Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$

$xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}$

$xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}$

$xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)$

$\implies \boxed{xyz = x + y + z + 2}$

(C)

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