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3 years ago

Which expression is equivalent to -6(-2/3+2x)? -4 -12x -4+2x 4 - 12x 4+12x

Mathematics

2 answers:

almond37 [142]3 years ago

7 0

Answer:

4-12x

Step-by-step explanation:

-2/3*-6 is 4

-6*2 is -12

4-12x

leva [86]3 years ago

5 0

Answer:

The third one, 4-12x

Step-by-step explanation:

You want to find an equation that equals 4 - 12 x

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Area for rectangle in the coordinate plane. IXL Geometry help pls !

Sidana [21]

Answer:

Step-by-step explanation:

The co-ordinates of the rectangle are:

A (8, 6)

B (-4, -10)

C (-8, -7)

D (4, 9)

The Area of Rectangle is Given by: <em>length x breadth </em>= AB x BC

Lengths of AB and BC can be found by distance formula

<em>length</em>(AB) = $\sqrt{(x_{2} - x_{1} )^{2} + (y_{2} - y_{1} )^{2}}$ = $\sqrt{(-4-8)^{2} + (-10-6)^{2}} = 20$

Similarly,

<em>length</em>(BC) = 5

Area of Rectangle = AB x BC = 20 x 5 = 100 square units

5 0

3 years ago

Use the strategy to simplify 4\sqrt(576)<br> Write the prime factorization of the radicand.

Ronch [10]

<h3>Answer: 96</h3>

===================================================

Work Shown:

$4\sqrt{576}\\\\4\sqrt{2*2*2*2*2*2*3*3}\\\\4\sqrt{(2*2)*(2*2)*(2*2)*(3*3)}\\\\4\sqrt{(2)^2*(2)^2*(2)^2*(3)^2}\\\\4\sqrt{(2*2*2*3)^2}\\\\4\sqrt{24^2}\\\\4*24\\\\96\\\\$

6 0

3 years ago

Read 2 more answers

Please help me and explain thank you so much I appreciate it

andreev551 [17]

Answer:

He gets 32, $10 bills.

Step-by-step explanation:

He puts in $630.

3 times $50 equals $150.

8 times $20 equals $160.

$150 plus $160 equals $310.

$630 minus $310 equals $320.

$320 divided by 10 equals 32.

So he gets 32, $10 bills.

8 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago

Y'=<img src="https://tex.z-dn.net/?f=%5Cfrac%7B-2x%2B2y%2B1%7D%7Bx-y%2B1%7D" id="TexFormula1" title="\frac{-2x+2y+1}{x-y+1}" alt

AlladinOne [14]

Answer:

see below

Step-by-step explanation:

dx 2+2xy 1 1 2 2

____ = ______ – ___x^2 – ____y^2 – _____x + ______y

dy 3 3 3 3 3

8 0

3 years ago