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3 years ago

Which direction does the graph of the equation shown below open? Y^2-4x+4y-4=0

Mathematics

2 answers:

zavuch27 [327]3 years ago

8 0

We have that
Y²<span>-4x+4y-4=0
</span>you can rewrite the equation as
4x=y²+4y-4
x=(y²/4)+y-1
x=0.25*(y²+4y)-1
x=0.25*(y²+4y+4-4)-1
x=0.25*(y²+4y+4)-5
x=0.25*(y+2)²-5

the equation of the parabola is of the form
x=a(y-k)²+h
is a horizontal parabola
where
(h,k)-----> is the vertex-----> (-5,-2)
a=0.25
a> 0--------> open to the right

the answer is
the option <span>D. Right
</span>
see the attached figure

Taya2010 [7]3 years ago

3 0

The answer is D, right because you rewrite the equation to a x= equation. Just by doing that, you know the equation has to be left or right. When rewriting the equation, the slope is positive, therefore it is facing right

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How many 6-digit numbers can be created using8, 0, 1, 3, 7, and 5 if each number is used only once?

Ludmilka [50]

Answer:

600 numbers

Step-by-step explanation:

For six-digit numbers, we need to use all digits 8,0,1,3,7,5 each once.

However, 0 cannot be used as the first digit, because it would make a 5-digit number.

Therefore

there are 5 choices for the first digit (exclude 0)

there are 5 choices for the first digit (include 0)

there are 4 choices for the first digit

there are 3 choices for the first digit

there are 2 choices for the first digit

there are 1 choices for the first digit

for a total of 5*5*4*3*2*1 = 600 numbers

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2 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

Which equation can be solved to find one of the missing side lengths in the triangle?

tester [92]

Answer:

cos(60) = 12

Step-by-step explanation:

I don't understand the question, but that is the only logical answer choice.

6 0

3 years ago

Explain whether the example is proportional or non-proportional.

kifflom [539]

Answer:

Porpotional

Step-by-step explanation:

because it is non proportional

5 0

3 years ago

Read 2 more answers

I need help! please!

mamaluj [8]

Answer:

$- \frac{1}{3}$

Step-by-step explanation:

The only thing important in this question in the original equation is the slope, which is 3, the multiplyer for x. To find the opposite slope follow the steps below.

The first step is to make the original slope into a fraction.

$\frac{3}{1}$

Then flip the numerator and denominator.

$\frac{1}{3}$

Now multiply this value by -1.

$\frac{1}{3} \times ( - 1) = - \frac{1}{3}$

There is the opposite slope of 3.

7 0

2 years ago