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Anton [14]
2 years ago
5

Can someone help me please

Mathematics
1 answer:
Margarita [4]2 years ago
7 0

Answer:

46

Step-by-step explanation:

Solution :

Remember that the sum of complementary angles is always 90°.

First, finding the value of x :

Set up an equation :

( Being complementary angles )

Solve for x

{ Remove unnecessary parentheses }

{ Combine like terms }

{ Subtract 3 from 30 }

{ Move 27 to right hand side and change it's sign }

{ Subtract 27 from 90}

{ Divide both sides by 9 }

The value of X is 7°

Now, Replacing the value of x in order to find the value of  B

{ Plug the value of x }

{ Multiply 7 by 7 }

{ Subtract 3 from 49 }

The measure of  B is 46°

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Find the missing terms.<br> 3, _____, _____, 21 (arithmetic)
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Answer:

Hello There!!

Step-by-step explanation:

The answer is 3,9,15,21.

hope this helps,have a great day!!

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Someone please be awesome and help me please :(
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Answer:

(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0

(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}

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Step-by-step explanation:

x^2+\frac{b}{a}x+\frac{c}{a}=0

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared.  Whatever you add in, you must take out.

x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0

Now we are read to write that one part (the first three terms together) as a square:

(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0

I don't see this but what happens if we find a common denominator for those 2 terms after the square.  (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0

They put it in ( )

(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0

I'm going to go ahead and combine those fractions now:

(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}

x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}

Now subtract b/(2a) on both sides:

x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}

Combine the fractions (they have the same denominator):

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

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sergij07 [2.7K]

Resolviendo el sistema de ecuaciones veremos que:

  • niña = 23kg
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  • perro = 18kg.

<h3>¿Como resolver el sistema de ecuaciones?</h3>

Aqui tenemos el sistema de ecuaciones:

Niña + niño = 51kg

Niño + perro = 46 kg

Niña + perro = 41kg

Para resolver esto, lo primero que debemos hacer es aislar una variable en una de las ecuaciones, por ejemplo, podriamos aislar "perro" en la tercera:

perro = 41kg - niña

Ahora reemplazamos eso en la segunda para obtener:

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Ahora logramos obtener la variable "niño" en terminos de la variable "niña". Podemos reemplazar esto en la primera ecuacion del sistema.

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niña = 46kg/2 = 23kg.

Ahora que sabemos esto, usamos las otras ecuaciones para encontrar el peso del niño y el perro:

niño = niña + 5kg = 23kg + 5kg = 28kg

perro = 41kg - niña = 41kg - 23kg = 18kg.

Sí quieres aprender más sobre sistemas de ecuaciones, puedes leer:

brainly.com/question/17174746

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