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2 years ago

11

Graph the line y=-3x -1

1 answer:

DochEvi [55]2 years ago

7 0

Answer: the graph line for y=-3x-1 is

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What is the slope intercept form of the linear equation that passes through the point (0, 9) and has a slope of -6?

STALIN [3.7K]

Answer:

y - 9 = -6( x - 0)

y - 9 = -6x + 0

y = -6x + 9

Step-by-step explanation:

7 0

3 years ago

Find the value of Z.<br> A)7<br> B)9<br> C)11<br> D)12

Gala2k [10]

As per the isosceles triangle theorem, if the two base angles are congruent then the legs are also congruent, so we must set the two legs equal to each other:

2z - 15 = 9

Add 15 to both sides:

15 - 15 = 0
15 + 9 = 24

Divide 2 from each side:

2z = 24

2z/2 = z

24/2 = 12

z = 12

Hence, the answer would be D: z = 12

4 0

2 years ago

Read 2 more answers

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

2 years ago

Tom is 10 years younger than Becky. In two years, Becky will be twice as old as Tom. How old is Tom now?

qwelly [4]

Answer:

20

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Rina8888 [55]

Answer:

Step-by-step explanation:

OK you should really have these by now.

Translation, reflection and rotations do not change segment lengths or angular relationships. They only change line slopes and/or y-intercepts.

4 0

2 years ago