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gtnhenbr [62]
3 years ago
13

How do you solve x/9-1/3=-5/3

Mathematics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

x = ( - 18)

Step-by-step explanation:

\frac{x}{9}  -  \frac{1}{3} = ( -  \frac{5}{3}  ) \\  \frac{x}{9} =  -  \frac{5}{3}   -  \frac{1}{3}  \\  \frac{x}{9}  =  -  \frac{6}{3}  \\  \frac{x}{9}  =  - 2 \\ 9 \times  \frac{x}{9}  =  - 2 \times 9 \\ x =  - 18

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aleksandr82 [10.1K]

Answer:

91

Step-by-step explanation:

6 0
3 years ago
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Uma barraca de tiro ao alvo de um parque de diversão dará um prêmio de r$ 20ao participante cada vez que ele acertar o alvo por
zavuch27 [327]

Resposta: 30

Explicação passo a passo:

Dado o seguinte:

Preço para atingir a meta = $ 20

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Número de vezes que o participante atinge o alvo.

Deixe o número de erros = m

Deixe o número de acertos = h

h + m = 80 - - - - (1)

20h - 10m = 100 - - - (2)

h = 80 - m

Substitua h = 80 - m em (2)

20 (80 - m) - 10m = 100

1600 - 20m - 10m = 100

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m = 50

De (h = 80 - m)

h = 80 - 50

h = 30

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O participante acerta o alvo 30 vezes

4 0
3 years ago
Review the proof.
andrezito [222]

Answer:

Step-by-step explanation:

A 2-column table with 8 rows. Column 1 is labeled Step with entries 1, 2, 3, 4, 5, 6, 7, 8. Column 2 is labeled Statement with entries cosine (2 x) = 1 minus 2 sine squared (x), let 2 x = theta, then x = StartFraction theta Over 2 EndFraction, cosine (theta) = 1 minus 2 sine squared (StartFraction theta Over 2 EndFraction), negative 1 + cosine (theta) = negative 2 sine squared (StartFraction theta Over 2 EndFraction), 1 + cosine (theta) = 2 sine squared (StartFraction theta Over 2 EndFraction), StartFraction 1 minus cosine (theta) Over 2 EndFraction = sine squared (StartFraction theta Over 2 EndFraction), sine (StartFraction theta Over 2 EndFraction) = plus-or-minus StartRoot StartFraction 1 minus cosine (theta) Over 2 EndFraction EndRoot.

Which step contains an error?

Answer:

 

Step-by-step explanation:

 

7 0
3 years ago
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A triangle has sides of 2,3 and 4 using the law of cosines , a^2+b^2-2abcosC=c^2 what is the value of 2abcosC
Anna007 [38]
\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\qquad 
\begin{cases}
a=2\\
b=3\\
c=4
\end{cases}
\\\\\\
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7 0
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jeka57 [31]

Answer:

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Step-by-step explanation:

n = 1  

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1.75 (DOUBLE) = 3.5

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8 0
3 years ago
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