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3 years ago

Cuánto es 120 por 12

Mathematics

1 answer:

Novay_Z [31]3 years ago

7 0

Answer:

120 x 12 = 1440

Step-by-step explanation:

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If 3a + 2b = 7 and 2a + 2b = 9 , what is the value of

Simora [160]

Hey there :)

We have two equations:
3a + 2b = 7
2a + 2b = 9

We need to solve simultaneously to find the values of a and b

eq.1 3a + 2b = 7
eq.2 ( 2a + 2b = 9 ) x -1 ) multiply by -1 to cancel 2b

3a + 2b = 7
- 2a - 2b = -9 ( Add both together )
-------------------
a = - 2

Substitute the value you found for a in a in order to find b

3( - 2 ) + 2b = 7 2( - 2 ) + 2b = 9
- 6 + 2b = 7 OR - 4 + 2b = 9
2b = 13 2b = 13
b = $\frac{13}{2}$ b = $\frac{13}{2}$

3 0

3 years ago

22,338 rounded to the nearest hundred

kogti [31]

22300 because the tenths place is below five

8 0

3 years ago

Read 2 more answers

A business owner opens one store in town A. The equation p(x)= 10,000(1.075)^ 5 represents the anticipated profit after t years.

lesya692 [45]

Answer:

It's 96%

Step-by-step explanation:

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2 years ago

Which of these represents categorical data?

Viktor [21]

Answer:

the answer is D) the most popular foods in the cafeteria

~batmans wife dun dun dun....

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3 years ago

Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).

I am Lyosha [343]

Take the homogeneous part and find the roots to the characteristic equation:

$y''+4y=0\implies r^2+4=0\implies r=\pm2i$

This means the characteristic solution is $y_c=C_1\cos2x+C_2\sin2x$ .

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form $y_p=ax\cos2x+bx\sin2x$ . Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With $y_1=\cos2x$ and $y_2=\sin2x$ , you're looking for a particular solution of the form $y_p=u_1y_1+u_2y_2$ . The functions $u_i$ satisfy

$u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$

where $W(y_1,y_2)$ is the Wronskian determinant of the two characteristic solutions.

$W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2$

So you have

$u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx$
$u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x$

$u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx$
$u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x$

So you end up with a solution

$u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

but since $\cos2x$ is already accounted for in the characteristic solution, the particular solution is then

$y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x$

so that the general solution is

$y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

7 0

3 years ago

Cuánto es 120 por 12​

Cuánto es 120 por 12