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Lelu [443]
3 years ago
14

Select the interval where g(x) < 0

Mathematics
1 answer:
natima [27]3 years ago
7 0

Answer:

4

Step-by-step explanation:

4 us the rifht answer just put 4

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Al llegar al hotel nos an dado un mapa con los lugares de la ciudad nos dijieron que 5 cm del mapa que representaban 600 metros
Sati [7]

Pregunta completa:

Al llegar al hotel nos an dado un mapa con los lugares de la ciudad nos dijieron que 5 cm del mapa que representaban 600 metros de la realidad hoy queremos ir a un par que que se encuentra a 8 cm del hotel en el mapa. ¿A qué distancia del hotel está la pareja?

Respuesta: 960m

Explicación paso a paso:

Dado que:

5 cm en el mapa equivale a 600 m en tierra Por lo tanto,

8 cm en el mapa será equivalente a:

5 cm en el mapa - - - - - -> 600 m en el suelo 8cm - - - - - - -> y metros en el suelo

Usando la multiplicación cruzada

y × 5 = 8 × 600

5y = 4800 Luego,

y = 4800/5

y = 960m

Por lo tanto, 8 cm en el mapa serán 960 m en realidad.

8 0
3 years ago
1. Solve for the area of the figure below.
wariber [46]

Answer:

39

Step-by-step explanation:

7 0
3 years ago
Please help with this question i will mark brainliest!
Blababa [14]

Answer:

hi im going to work on this i wiil comment answer in a minute

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A student solves the following equation and
Phoenix [80]

Answer:

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Step-by-step explanation:

Considering the expression

\frac{3}{a+2}-6\cdot \frac{a}{-4+a^2}=\frac{1}{a-2}

\frac{3}{a+2}-\frac{6a}{-4+a^2}=\frac{1}{a-2}

\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)

\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)

\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)

as

  • \frac{3}{a+2}\left(a+2\right)\left(a-2\right):\quad 3\left(a-2\right)
  • -\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right):\quad -6a
  • \frac{1}{a-2}\left(a+2\right)\left(a-2\right):\quad a+2

so equation becomes

3\left(a-2\right)-6a=a+2  

-3a-6=a+2

-3a-6+6=a+2+6

-4a=8

\mathrm{Divide\:both\:sides\:by\:}-4

\frac{-4a}{-4}=\frac{8}{-4}

a=-2

\mathrm{Verify\:Solutions}

\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}

\mathrm{Solve\:}\:a+2=0:\quad a=-2

\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2

\mathrm{Solve\:}\:a-2=0:\quad a=2

So the following points are undefined

a=-2,\:a=2

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

4 0
3 years ago
Josh was given $50 for his birthday He spent $5.52 on cancy S325 on books.
Readme [11.4K]

Answer:

$20.11 (assuming S325 was meant to be $3.25)

Step-by-step explanation:

50.00 - 5.52 = 44.48

44.48 - 3.25 = 41.23

41.23 -21.12 = 20.11

Hope this helps <3

3 0
3 years ago
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