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Alex_Xolod [135]
3 years ago
9

Find all x so that x^5=-x^3. Imaginary numbers (like "i") are allowed.

Mathematics
2 answers:
natita [175]3 years ago
8 0
It’s 3 keidiskdkkskskskkssiisisisisiis
MissTica3 years ago
5 0
Finding x like finding imaginary number I would say like in this problem x is more than 5 and x is less than 3 so 5x-3x
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Select three ratios that are equivalent to <br> 4:3 <br> Choose 3 answers:
postnew [5]
Ratios equivalent to 4:3 would be 8:6 16:12 and 24:18
6 0
3 years ago
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A coordinate plane with a line passing through the points (negative 4, negative 5) and (1, negative 1).
Ksju [112]

Answer:

4x - 5y - 9 = 0

\frac{9}{4}

\frac{-9}{5}

Step-by-step explanation:

The equation of a straight line passing through the points (-4,-5) and (1,-1) is given by \frac{y - (-5)}{-5-(-1)} =\frac{x-(-4)}{-4-1}

⇒ \frac{y+5}{-4} =\frac{x+4}{-5}

⇒ 5y + 25 = 4x + 16

⇒ 4x - 5y - 9 = 0 ....(1) (Answer)

Now, this above equation can be represented as

\frac{x}{\frac{9}{4} } + \frac{y}{\frac{-9}{5} } =1

Therefore, the x-intercept is \frac{9}{4} (Answer)

And the y-intercept is \frac{-9}{5}. (Answer)

7 0
3 years ago
Read 2 more answers
104x1033..........................and also should i quit fortnite?
Dimas [21]

Answer:

yes you should its a waste of time and money

6 0
3 years ago
Read 2 more answers
TRANSLATIONS ON THE COORDINATE PLANE
klio [65]

Answer:

idk

Step-by-step explanation:

idk

8 0
2 years ago
Sea un cuadrado de 2 pulgadas de lado uniendo los puntos medios se obtiene otro cuadrado inscrito en el anterior si repetimos es
Ne4ueva [31]

Answer:

1) La serie geométrica formada es

4, 2, 1,..., ∞

2) La suma al infinito de las áreas de los cuadrados es 8 in.²

Step-by-step explanation:

1) El área del primer cuadrado, a₁ = 2² = 4 pulgadas²

El área del siguiente cuadrado, a₂ = (√ (1² + 1²)) ² = (√2) ² = 2 pulg²

El área del siguiente cuadrado, a₃ = ((√ (2) / 2) ² + (√ (2) / 2) ²) = 1 pulg²

Por lo tanto, la razón común, r = a₂ / a₁ = 2/4 = a₃ / a₂ = 1/2

Las áreas de los cuadrados progresivos forman una progresión geométrica como sigue;

4, 4×(1/2), 4 ×(1/2)²,...,4×(1/2)^{\infty}

De donde obtenemos la serie geométrica formada de la siguiente manera;

4, 2, 1,..., ∞

2) La suma de 'n' términos de una progresión geométrica hasta el infinito para -1 <r <1 se da como sigue;

S_{\infty} = \dfrac{a}{1 - r}

Por lo tanto, la suma de las áreas de los cuadrados hasta el infinito se obtiene sustituyendo los valores de 'a' y 'r' en la ecuación anterior de la siguiente manera;

La \ suma \ al \ infinito \ del \ cuadrado \ S_{\infty}  = \dfrac{4 \ in.^2}{1 - \dfrac{1}{2} } = \dfrac{4 \ in.^2}{\left(\dfrac{1}{2} \right)} = 2 \times 4 \ in.^2= 8 \ in.^2

La suma al infinito de las áreas de los cuadrados, S_{\infty} = 8 in.²

7 0
2 years ago
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