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2 years ago

What is the equation

Mathematics

2 answers:

slava [35]2 years ago

8 0

Answer:

B. $(x+1)^2+(y-2)^2=9$

Step-by-step explanation:

The given circle is centered at $(-1,2)$ and has radius 3 units.

The equation is given by

$(x-a)^2+(y-b)^2=r^2$

where (a,b) is the center and r is the radius.

We substitute the values to obtain;

$(x--1)^2+(y-2)^2=3^2$

$(x+1)^2+(y-2)^2=9$

alexandr1967 [171]2 years ago

4 0

The answer is the letter B.

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5q greater than or equal to 8q -3/2

kakasveta [241]

5q ≥ 8q - 3/2

Add 3/2 to both sides.

3/2 + 5q ≥ 8q

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3/2 ≥ 3q

Multiply both sides by 2.

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Divide both sides by 6.

0.5 ≥ q.

The value of q is less than or equal to 0.5.

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2 years ago

15x+2y>20 i need to change this to slope intercept form, please show steps!!

aleksandrvk [35]

Answer:

y>-15/2x+10

Step-by-step explanation:

15x+2y>20

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3 years ago

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2 years ago

What is the value of b if 9+b=-8

kkurt [141]

9+B=-8 first start by moving the nine
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hope this helps and give brainliest . thanks !

5 0

2 years ago

Read 2 more answers

If a,b,c,d, and e are non negative such that x^6-289x^4-x^2+289=(x-a)(x+b)(x_c)(x+d)(x^2+ex+1), what is the value of a+b+c+d+e?

Irina18 [472]

Answer:

Step-by-step explanation:

Hello,

If $X=x^2$ it becomes

$X^3-289X^2-X+289=(X^2-1)(X-289)$

and $289=17^2$

$X^2-1=(X-1)(X+1)=(x^2-1)(x^2+1)=(x+1)(x-1)(x^2+1) \\\\x^2-289=x^2-17^2=(x-17)(x+17)$

So,

$x^6-289x^4-x^2+289=(x^2-1)(x^2+1)(x^2-289)=(x-1)(x+1)(x+17)(x-17)(x^2+1)$

a = 1

b = 1

c = 17

d = 17

e = 0

Then a + b + c + d + e = 36

Thanks

3 0

2 years ago

What is the equation​

What is the equation