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Korolek [52]
3 years ago
10

Square root of 15 please help me

Mathematics
2 answers:
VashaNatasha [74]3 years ago
8 0

Answer:

Decimal Form Answer for your Square root of 15

3.87298334

Step-by-step explanation:

Blababa [14]3 years ago
7 0

Answer:

approx 3.87298

Step-by-step explanation:

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En el estante de un negocio hay 2 tipos de tarros de la misma mermelada y marca. El tarro más alto tiene el doble de altura que
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Answer:

tarro corto

Step-by-step explanation:

Aquí tenemos que comprar el frasco que tiene el menor costo por volumen.

h_1 = Altura del frasco corto

h_2 = La altura del frasco alto es el doble que el del frasco corto. = 2h_1

d_1 = Diámetro del frasco corto

d_2 = El diámetro del frasco alto es la mitad del frasco corto = \dfrac{1}{2}d_1

El volumen de un cilindro es \pi \dfrac{d^2}{4}h

La razón de los volúmenes de los frascos es

\dfrac{V_1}{V_2}=\dfrac{\pi\dfrac{d_1^2}{4}h_1}{\pi\dfrac{d_2^2}{4}h_2}\\\Rightarrow \dfrac{V_1}{V_2}=\dfrac{d_1^2h_1}{d_2^2h_2}\\\Rightarrow \dfrac{V_1}{V_2}=\dfrac{d_1^2h_1}{(\dfrac{1}{2}d_1)^22h_1}\\\Rightarrow \dfrac{V_1}{V_2}=\dfrac{d_1^2h_1}{\dfrac{1}{4}d_1^22h_1}\\\Rightarrow \dfrac{V_1}{V_2}=\dfrac{1}{\dfrac{1}{2}}\\\Rightarrow \dfrac{V_1}{V_2}=2\\\Rightarrow V_1=2V_2

El costo del frasco corto por unidad de volumen es

\dfrac{8000}{V_1}=\dfrac{8000}{2V_2}=\dfrac{4000}{V_2}

El costo del frasco alto por unidad de volumen es

\dfrac{4500}{V_2}=\dfrac{4500}{V_2}

\dfrac{4000}{V_2}

Entonces, el costo del frasco corto por unidad de volumen es menor que el costo por unidad de volumen del frasco alto.

Por lo tanto, deberíamos tomar el frasco corto.

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3 years ago
Describe a situation that could be modeled with the ratio 3: 1.
ehidna [41]

Answer:

three boys to 1 girl

Step-by-step explanation:

a

5 0
3 years ago
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What is the step of it
andrew11 [14]
-9/3 + 4

-3 + 4 = 1

The answer to the question
7 0
3 years ago
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Somebody help me please fast
Nina [5.8K]

Answer:

768 cubic inches

Step-by-step explanation:

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3 years ago
Find a fraction equivalent to 5/7 whose squared terms add up to 1184.
bogdanovich [222]

The system of equations of two unknowns is formulated and solved.

\large\displaystyle\text{$\begin{gathered}\sf \bf{ \left\{\begin{matrix} \ \ \ \dfrac{x}{y} = \dfrac{5}{7} \\ x^2+y^2 = 1184 \end{matrix}\right. \ \Longrightarrow \ x=\dfrac{5}{7}y  } \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf \bf{ \left (\dfrac{5}{7}y \right )^2+y^2=1184\ \Longrightarrow\ 25y^2+49y^2=58016 } \end{gathered}$}

                                                              \large\displaystyle\text{$\begin{gathered}\sf \bf{74y^{2}=58016} \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf \bf{ \ \  \ \ \ \ y^{2}=784 } \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf \bf{ \ \ \ \ \ y=\pm\sqrt{784}=\pm28  } \end{gathered}$}

                                                                  \large\displaystyle\text{$\begin{gathered}\sf \bf{ x=\dfrac{5}{7}(\pm 28)=\pm 20  } \end{gathered}$}

The fraction that satisfies the request is \bf{\dfrac{20}{28}} , since in \bf{\dfrac{-20}{-28}} the negative signs are canceled and the first fraction is obtained.

3 0
2 years ago
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