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2 years ago

Can someone help me please

Mathematics

1 answer:

Margarita [4]2 years ago

7 0

Answer:

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. 3, _____, _____, 21 (arithmetic)

Rom4ik [11]

Answer:

Hello There!!

Step-by-step explanation:

The answer is 3,9,15,21.

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4 0

2 years ago

Read 2 more answers

Someone please be awesome and help me please :(

solong [7]

Answer:

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

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

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

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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3 years ago

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3 years ago

3/8+1/6 equals tokyxldlyx

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3 years ago

Read 2 more answers

¿cuanto pesa cada uno de nuestros 3 personajes?

sergij07 [2.7K]

Resolviendo el sistema de ecuaciones veremos que:

niña = 23kg
niño = 28kg
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:

niño + (41kg - niña) = 46kg

niño - niña = 46kg - 41kg = 5kg

niño = niña + 5kg

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

niña + niño = 51kg

niña + (niña + 5kg) = 51kg

2*niña = 51kg - 5kg = 46kg

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

6 0

2 years ago