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3 years ago

Solve x^2-12x+36 using the quadratic formula

Mathematics

2 answers:

Paladinen [302]3 years ago

6 0

Answer:

(x - 6) ${2}$

Step-by-step explanation:

Solve the equation for x by finding a, b, and c of the quadratic then applying the quadratic formula x = 6 Double roots

leonid [27]3 years ago

3 0

Answer:

x=6

Step-by-step explanation:

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Pls help I will mark brainlist if correct !! :)

Zielflug [23.3K]

Answer:

the first one is correct

Step-by-step explanation:

mark brainliest please

4 0

3 years ago

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Eric and Frank want to equally share 4/3 feet of rope. what length of rope should each friend get? explain how to use a drawing

blondinia [14]

The rope is 1 and 1/3 long,1 is the same as saying 3/3,
this means that the rope is 3/3 + 1/3 long,so it is 4/3 foot long,
you now want to get half of this number so that each person has an equal length of rope,so multiply the length of rope by a half,
4/3 x 1/2 = 4/6 , which if simplified = 2/3, this means that each friend gets 2/3 foot of rope.

You can draw this in a diagram by drawing the 1 foot of rope, and then dividing it up into 3 sections. Then add the 1/3 foot of rope into the drawing. It should then be easy to see that altogether there are 4 separate sections of rope and that if each friend has two sections then they will be sharing equal amounts, so consequently each friend has two of the thirds of rope, 2/3

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3 years ago

Write an equivalent expression 2x + 10

balu736 [363]

Answer: 2 (x+5)

Step-by-step explanation:

7 0

2 years ago

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What is the average rate of change <br>y - 3 = -12 (x + 4)

vagabundo [1.1K]

Answer:

$- 12x$

Step-by-step explanation:

$y - 3 = - 12(x + 4) \\ y - 3 = - 12x - 48 \\ y = - 12x - 45$

Thr answer is the slope.

4 0

3 years ago

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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