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3 years ago

9.9.9.9 in expanded notation. I need to know the value for it.

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

4 0

Answer:

9.9.9.9 = ( 9*1) . (9/10) + (0/100) + (9/1000) + (0/10000) +(9/100000)

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What is the image of (-9,3)(−9,3) after a reflection over the y-axis?

nirvana33 [79]

Answer:

(9, 3 )

Step-by-step explanation:

under a reflection in the y- axis

a point (x, y ) → (- x, y ) , then

(- 9, 3 ) → (9, 3 )

3 0

1 year ago

Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.

jok3333 [9.3K]

Answer:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

$\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}$

And, the distance from P to (0, -2) will be:

$\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}$

So sum of the two distances must be 6. Therefore:

$d_1+d_2=6$

Now, by substitution:

$(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6$

Simplify. We can subtract the second term from the left:

$\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}$

Square both sides:

$(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)$

We can cancel the x² terms and continue squaring:

$y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4$

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

$-8y=36-12\sqrt{x^2+(y+2)^2}$

We can divide both sides by -4:

$2y=-9+3\sqrt{x^2+(y+2)^2}$

Adding 9 to both sides yields:

$2y+9=3\sqrt{x^2+(y+2)^2}$

And, we will square both sides one final time.

$4y^2+36y+81=9(x^2+(y^2+4y+4))$

Distribute:

$4y^2+36y+81=9x^2+9y^2+36y+36$

The 36y will cancel. So:

$4y^2+81=9x^2+9y^2+36$

Subtracting 4y² and 36 from both sides yields:

$9x^2+5y^2=45$

And dividing both sides by 45 produces:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

5 0

3 years ago

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1/3 of 9 is 3<br> How can you figure out these problems

gtnhenbr [62]

I don’t see the problems you want me to solve. Sorry

6 0

2 years ago

Please help xx 10 points

zvonat [6]

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bit. $^{}$ ly/3a8Nt8n

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2 years ago

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Find the value of x. If necessary, write your answer in the simplest radical form.

Paul [167]

Answer:

x = 4 $\sqrt{5}$

Step-by-step explanation:

Using Pythagoras' identity in the right triangle.

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

x² + 1² = 9²

x² + 1 = 81 ( subtract 1 from both sides )

x² = 80 ( take square root of both sides )

x = $\sqrt{80}$ = $\sqrt{16(5)}$ = $\sqrt{16}$ × $\sqrt{5}$ = 4 $\sqrt{5}$

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2 years ago

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