MATHEMATICS MIDDLE SCHOOL

What is the radius of the circle given the equation x 2 - 6x + y 2 + 10y - 2 = 0?

Answers

Answer 1

Answer:

we are given

$x^2-6x+y^2+10y-2=0$

Firstly, we will change it into standard equation of circle

that is

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

where r is radius

(h,k) is center

now, we can change our equation into this form

To change our equation into this form, we need to complete x square and y square

$x^2-6x+y^2+10y-2=0$

step-1: Move 2 on right side

$x^2-6x+y^2+10y-2+2=0+2$

$x^2-6x+y^2+10y=2$

step-2: Complete x square

$x^2-2*3*x+y^2+10y=2$

$x^2-2*3*x+3^2+y^2+10y=2+3^2$

$(x-3)^2+y^2+10y=2+3^2$

step-3: Complete y square

$(x-3)^2+y^2+2*5*y=2+3^2$

$(x-3)^2+y^2+2*5*y+5^2=2+3^2+5^2$

$(x-3)^2+(y+5)^2=2+3^2+5^2$

step-4: Combine right side terms

$(x-3)^2+(y+5)^2=36$

$(x-3)^2+(y+5)^2=6^2$

now, we can compare with our equation

and we will get

radius =r=6...............Answer

center=(h,k)=(3,-5)........................Answer

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If i ate 10 cookies in a 8 day period of time how many cookies will i eat in a 12 day period

Answers

10 cookies in a 8 day period
? cookies in a 12 day period

10 cookies in a 8 day period

Divide both numbers by 2

5 cookies in a 4 day period

Multiply both numbers by 3

15 cookies in a 12 day period.

You will eat 15 cookies in a 12 day period.

Alice wants to find all the prime factors of the number you get when you multiply 17 x 11 x 13 x 7. She thinks she has to use a calculator to perform all the multiplications and then find the prime factorization of the resulting number. Do you agree? Why or why not?

Answers

I disagree because she can use a pencil and paper or she can do mental math

I really need help!!! It would mean a lot if someone could help me.

Answers

The y answers would be 11, 8, 5, 2, -1.

To find these parts of the table, we simply put the x value in for x and solve for y. The first two are done for you below.

WHEN x = -2

y = -3x + 5

y = -3(-2) + 5

y = 6 + 5

y = 11

WHEN x = -1

y = -3x + 5

y = -3(-1) + 5

y = 3 + 5

y = 8

-3(-2) +5=11

-3(-1) +5=8

-3(0) +5=5

-3(1) +5=2

-3(2) +5=-1

Find the sum (9x+4) + (6x-6)

Answers

Hey mate !!

Here's the answer !!

All the terms in the question are like terms. Hence we can add them as it is. Refer to the attachment for the answer !!

Hope it helps !!

Cheers !!

9x + 4) + (6x - 6)
15x - 2 = 0
15x - 2 + 2 = 2
15x = 2
X = 2/15 = 0.133333333

The edge of one cube is 4 m shorter than the edge of a second cube. The volumes of the two cubes differ by 1216 m^3. Find the edge of the smaller cube.

Answers

Volume of cube, V = edge^3
Let edge of cube#1 = (x-4) m, therefore volume of cube#1, v1 = (x-4)^3 m
Let edge of cube#2 = x m, therefore volume of cube#2, v2 = x^3 m
Diff. in volume (in m) = 1216 = v2-v1 = [ x^3 - (x-4)^3 ]
= x^3 - [(x-4)(x-4)(x-4)]
= x^3 - [x^2 - 8x +16(x - 4)]
= x^3 - [ x^3 - 12x^2 + 48x - 64 ]
= 12x^2 - 48x + 64
= 4 (3x^2 - 12x + 16)
Therefore 4 (3^2 - 12x + 16) = 1216
3x^2 - 12x + 16 = 1216/4 = 304
3x^2 - 12x - 288 = 0
3 (x^2 - 4x - 96) = 0
(x^2 - 4x - 96) = 0
(x - 12) (x + 8) =0
(x-12) = 0
Therefore x = 12 m
Edge of cube#2 = x m = 12m
Edge of cube#1 = (x-4) m = 8m

$a- the\ edge\ of\ the\ smaller\ cube\ \ \ \wedge\ \ \ a>0\nV_a-the\ volume\ of\ the\ smaller\ cube\ \ \ \Rightarrow\ \ \ V_a=a^3\n \n(a+4)^3-a^3=1216\ \ \ \wedge\ \ \ (x+y)^3=x^3+3x^2\cdot y+3x\cdot y^2+y^3\n \na^3+3a^2\cdot 4+3a\cdot 4^2+4^3-a^3=1216\n \n12a^2+48a+64-1216=0\ \ \ \Rightarrow\ \ \ 12a^2+48a-1152=0\ /:12\n \na^2-4a-96=0\ \ \ \Rightarrow\ \ \ \Delta=(-4)^2-4\cdot 1\cdot(-96)=16+384=400\n \n √(\Delta) =20\n \na_1= (4-20)/(2\cdot1) =-8<0,\ \ \ a_2= (4+20)/(2\cdot1) =12>0$

$Ans.\ The\ edge\ of\ the\ smaller\ cube=12\ m$

Miles has a train collection with 36 engines 72 boxcars and 18 cabooses he wants to arranged the train cars in equal rows with only one type of train car in each row How many can he put in each row.

Answers

In order to find how many trains can be put in each row while having only one type of car in each row, you must find the GCF of all the numbers of cars.

First factor all three numbers
36 Engines= 2*2*3*3
72 boxcars = 2*2*2*3*3
18 cabooses= 2*3*3

Since each number contains at least 2*3*3, or 18, each row can be up to 18 train cars in length.
In this case, there will be 1 row of cabooses,
18 ÷18=1
2 rows of engines,
36÷18=2
and 4 rows of boxcars
72÷18=4

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