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3 years ago

La cuenta de la luz en enero fue de 35600 y si en febrero la cuenta baja en un 12% cuál es el monto que debe pagar en febrero

Mathematics

1 answer:

Lapatulllka [165]3 years ago

7 0

Answer:

¿Cuánto paga tu vecino por la electricidad? Si toma la cantidad promedio de electricidad consumida por mes y la multiplica por x 911) verá que el costo residencial promedio de la electricidad en el mes de enero indica que las ventas totales de electricidad cayeron 4.9% en el primer semestre de 2016

Step-by-step explanation:

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HELP ME QUICK: Suppose AB has one endpoint at A(0, 0). If (5, 3) is the midpoint of AB, what are the coordinates of point B?

DiKsa [7]

Answer:

Step-by-step explanation:

If P is midpoint of AB and: $A(0,\,0)\,,\quad P(5,\,3)\,,\quad B(x_B,\,y_B)$

then:

$x_P-x_A=x_B - x_P\qquad\quad\ \wedge\qquad y_P-y_A=y_B - y_P\\\\ 5-0=x_B-5\qquad\quad\wedge\qquad 3-0=y_B -3\\\\ x_B=5+5\qquad\qquad\wedge\qquad\ \ y_B=3+3 \\\\ {}\quad x_B=10\qquad\qquad\wedge\qquad\qquad \ y_B=6$

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3 years ago

Sofia is helping set up for a wedding. There are 80 chairs that need to be arranged in a room, and 60% of those chairs are still

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48 is the answer of your mathematics question on this site/app.

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3 years ago

If a student solves 5 problems in 10 minutes what is the proportional constant

drek231 [11]

5/10=2
So 2 minutes per problem

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3 years ago

Read 2 more answers

Find the value of each variable.<br> (y + x)<br> 2xx(y- x)

mart [117]

X=0 and y=0, set them equal to each other

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4 years ago

Solve this differential equation using power series and indicial roots about (0,0):

Ivanshal [37]

Let $y=\displaystyle\sum_{k\ge0}a_kx^k$ , so that

$y'=\displaystyle\sum_{k\ge1}ka_kx^{k-1}$
$y''=\displaystyle\sum_{k\ge2}k(k-1)a_kx^{k-2}$

Substituting into the ODE gives

$\displaystyle3x\sum_{k\ge2}k(k-1)a_kx^{k-2}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0$
$\displaystyle3\sum_{k\ge2}k(k-1)a_kx^{k-1}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0$

The first series starts with a linear term, while the other two start with a constant. Extract the first term from each of the latter two series:

$\displaystyle6\sum_{k\ge1}ka_kx^{k-1}=6\sum_{k\ge2}ka_kx^{k-1}+6a_1$
$\displaystyle\sum_{k\ge0}a_kx^k=\sum_{k\ge1}a_kx^k+a_0$

Finally, to get the series to start at the same index, shift the index of the first two series by replacing $k$ with $k+1$ . Then the ODE becomes

$\displaystyle3\sum_{k\ge1}k(k+1)a_{k+1}x^k+6\sum_{k\ge1}(k+1)a_{k+1}x^k+\sum_{k\ge1}a_kx^k+6a_1+a_0=0$

which can be consolidated to get

$\displaystyle\sum_{k\ge1}\bigg[(3k(k+1)+6(k+1))a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0$
$\displaystyle\sum_{k\ge1}\bigg[3(k+1)(k+2)a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0$

You're fixing the solution so that it contains the origin, which means

$y(0)=\displaystyle\sum_{k\ge0}a_kx^k=a_0=0$

which in turn means $a_1=0$ . With the given recurrence, it follows that $a_k=0$ for all $k\ge2$ , so the solution would be $y=0$ . This is to be expected, since $x=0$ is clearly a singular point for the ODE.

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4 years ago