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docker41 [41]
3 years ago
8

17

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
3 0

Answer:

Center =  \boxed{1}, \: \boxed{ - 2} \: ; \: Radius =  \boxed{3}

Step-by-step explanation:

{x}^{2}  +  {y}^{2}  - 2x + 4y - 4 = 0 \\  \\  ({x}^{2}  - 2x + 1 - 1) + ( {y}^{2}  + 4y + 4 - 4) - 4 = 0 \\  \\ ( {x}^{2}  - 2x + 1) + ( {y}^{2}  + 4y + 4) - 9 = 0 \\  \\  {(x - 1)}^{2}  +  {(y + 2)}^{2}  = 9 \\  \\ {(x - 1)}^{2}  +  {(y + 2)}^{2}  =  {3}^{2}  \\  \\ equating \: it \: with \\  \\  {(x - h)}^{2}  +  {(y - k)}^{2}  =  {r}^{2}  \\  \\ h = 1 \\ k =  - 2 \\ r = 3 \\  \\ center =  \boxed{1} \: \boxed{ - 2} \:  \: radius =  \boxed{3}

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The slope of the line that passes through the points (-8,-7) and (-4,-2)
goldfiish [28.3K]

Answer:

I'm pretty sure the slope is 5/4

Step-by-step explanation:

When you graph (-8,-7) and (-4,-2) you use rise/run to find the slope.Which is essntially counting up and the over from the first point to the second. Or you can use the equation of slope. Which is y1-y2/x1-x2.

7 0
3 years ago
What is 3.42 as a fraction and what is it simplified
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3.42  = 3 42/100

divide the numerator and denominator by 2

3 21/50


4 0
3 years ago
Read 2 more answers
Area and perimeter please
stealth61 [152]

Answer:

185.68 for perimeter

1652.16 for area

Step-by-step explanation:

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5 0
4 years ago
Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

-6=b

Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

7 0
3 years ago
Joanne wants to put wallpaper in her daughter's room. Each wall measure 16.8 feet long by 12.9 feet wide. What is the area of on
Leokris [45]
The wall would be 216.72 feet for Joanna’s wall.
8 0
3 years ago
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