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3 years ago

5

At a shoe store there are 40 pairs of shoes. 8 pairs are yeezys.15 pairs of all shoes have velcro and 2 yeezys have velcro. How

many pairs of shoes are not yeezys and do not have velcro

2 answers:

shusha [124]3 years ago

8 0

Answer:

40 - 15 - 2= 23 we simply do not count the yezzys until they mention velcro and just takeaway 15 pairs of shoes and 2 velcro, yezzy that have velcro

are 2 so it must be 23.

Step-by-step explanation:

Gemiola [76]3 years ago

4 0

Answer:

17

Step-by-step explanation:

add the number of shoes with velcro, not including the 2 yeezys because their already included with the total number of yeezys, to the number of yeezys

15 + 8 = 23

subtract 23 from 40

40 - 23 = 17

17 of the shoes aren't yeezys and do not have velcro

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<h3>Factorization</h3>

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2 years ago

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

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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:

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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:

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

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

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:

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

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

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:

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3 years ago

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3 years ago

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Answer:

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