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2 years ago

HELP!!! PLEASE HURRY!!!!

Mathematics

2 answers:

yulyashka [42]2 years ago

7 0

Answer:

B, I solved then used interval notion to convert it.

nydimaria [60]2 years ago

5 0

Answer:

the answer is b

Answer:

the answer is b

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On a game show, a contestant randomly chooses a chip from a bag that contains numbers and strikes. The theoretical probability o

notka56 [123]

If the probability of choosing a strike is 3/10, that means that from the total items available 3 out of 10 are strikes, now, if there are 9 strikes total, this is the equation representing such result:
(3/10)(totalChips) = 9
if we solve for totalChips we have:
3(totalChips) = 9*10 = 90
totalChips = 90/3 = 30
therefore the total amount of chips in the bag is 30

4 0

3 years ago

Read 2 more answers

What are 3 ratios that are equivalent to 8 :5

Andrei [34K]

Answer:

Step-by-step explanation:

8/5 = 16/10 = 24/15

8:5 = 16:10 = 24:15

8 0

3 years ago

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. Match the circle equations in general form with t

ArbitrLikvidat [17]

Answer:

1) For $x^2 + y^2 - 4x + 12y - 20 = 0$ , the standard form is $(x-2)^2 + (y+6)^2 = 60\\$

2) For $x^2 + y^2 + 6x - 8y - 10 = 0$ , the standard form is $(x + 3)^2 + (y - 4)^2 = 35\\$

3) For $3x^2 + 3y^2 + 12x + 18y - 15 = 0$ , the standard form is $(x + 2)^2 + (y+ 3)^2 = 18\\$

4) For $5x^2 + 5y^2 - 10x + 20y - 30 = 0$ , the standard form is $(x - 1)^2 + (y+ 2)^2 = 11\\$

5) For $2x^2 + 2y^2 - 24x - 16y - 8 = 0$ , the standard form is $(x - 6)^2 + (y+ 4)^2 = 56\\$

6) For $x^2 + y^2 + 2x - 12y - 9 = 0$ , the standard form is $(x+1)^2 + (y-6)^2 = 46\\\\$

Step-by-step explanation:

This can be done using the completing the square method.

The standard equation of a circle is given by $(x - a)^2 + (y-b)^2 = r^2$

1) For $x^2 + y^2 - 4x + 12y - 20 = 0$

$x^2 - 4x + y^2 + 12y = 20\\x^2 - 4x + 2^2 + y^2 + 12y + 6^2 = 20 + 4 + 36\\(x-2)^2 + (y+6)^2 = 60\\$

Therefore, for $x^2 + y^2 - 4x + 12y - 20 = 0$ , the standard form is $(x-2)^2 + (y+6)^2 = 60\\$

2) For $x^2 + y^2 + 6x - 8y - 10 = 0$

$x^2 + 6x + y^2 - 8y = 10\\x^2 + 6x + 3^2 + y^2 - 8y + 4^2 = 10 + 9 + 16\\(x + 3)^2 + (y- 4)^2 = 35\\$

Therefore, for $x^2 + y^2 + 6x - 8y - 10 = 0$ , the standard form is $(x + 3)^2 + (y - 4)^2 = 35\\$

3) For $3x^2 + 3y^2 + 12x + 18y - 15 = 0$

Divide through by 3

$x^2 + y^2 + 4x + 6y = 5$

$x^2 + y^2 + 4x + 6y = 5\\x^2 + 4x + 2^2 + y^2 + 6y + 3^2 = 5 + 4 + 9\\(x + 2)^2 + (y+ 3)^2 = 18\\$

Therefore, for $3x^2 + 3y^2 + 12x + 18y - 15 = 0$ , the standard form is $(x + 2)^2 + (y+ 3)^2 = 18\\$

4) For $5x^2 + 5y^2 - 10x + 20y - 30 = 0$

Divide through by 5

$x^2 + y^2 - 2x + 4y = 6$

$x^2 + y^2 -2x + 4y = 6\\x^2 - 2x + 1^2 + y^2 + 4y + 2^2 = 6 + 1 + 4\\(x - 1)^2 + (y+ 2)^2 = 11\\$

Therefore, for $5x^2 + 5y^2 - 10x + 20y - 30 = 0$ , the standard form is $(x - 1)^2 + (y+ 2)^2 = 11\\$

5) For $2x^2 + 2y^2 - 24x - 16y - 8 = 0$

Divide through by 2

$x^2 + y^2 - 12x - 8y = 4$

$x^2 + y^2 - 12x - 8y = 4\\x^2 - 12x + 6^2 + y^2 - 8y + 4^2 = 4 + 36 + 16\\(x - 6)^2 + (y+ 4)^2 = 56\\$

Therefore, for $2x^2 + 2y^2 - 24x - 16y - 8 = 0$ , the standard form is $(x - 6)^2 + (y+ 4)^2 = 56\\$

6) For $x^2 + y^2 + 2x - 12y - 9 = 0$

$x^2 + 2x + y^2 - 12y = 9\\x^2 + 2x + 1^2 + y^2 - 12y + 6^2 = 9 + 1 + 36\\(x+1)^2 + (y-6)^2 = 46\\$

Therefore, for $x^2 + y^2 + 2x - 12y - 9 = 0$ , the standard form is $(x+1)^2 + (y-6)^2 = 46\\\\$

8 0

3 years ago

How do you solve <br> y= 1/8 x+3

cupoosta [38]

Answer:

Leave it

Step-by-step explanation:

1/8x has a different variable than 3 so the only way to solve it is to leave it the same

5 0

3 years ago

Read 2 more answers

By using the equation below complete the square, write the equation of the circle in standard form from the following equation.

Lubov Fominskaja [6]

Given:

The equation of a circle is

$10y+x^2=-18+5x-y^2$

To find:

The center and radius of the given equation by completing the square.

Solution:

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$ ...(i)

where, (h,k) is center and r is radius of the circle.

We have,

$10y+x^2=-18+5x-y^2$

It can be written as

$(x^2-5x)+(y^2+10y)=-18$

$\left(x^2-5x+\left(\dfrac{5}{2}\right)^2\right)+\left(y^2+10y+\left(\dfrac{10}{2}\right)^2\right)=-18+\left(\dfrac{5}{2}\right)^2+\left(\dfrac{10}{2}\right)^2$