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3 years ago

Can you solve -4a^ - (6a-2) + 3 (a^-1).the arrows that are pointing up is exponent 2.also please show work

Mathematics

1 answer:

Burka [1]3 years ago

5 0

First step is to distribute.
-4a^2 - 6a +2 + 3a^2 -3
Combine like terms
-a^2 -6a -1

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Answer:

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Answer:

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3 years ago

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3 years ago

Write the equation of circle b with center B(-2,3) that passes through (1,2)

professor190 [17]

Answer:

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

Step-by-step explanation:

The standard equation for circle is

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

where point (a,b) is coordinate of center of circle and r is the radius.

______________________________________________________

Given

center of circle =((-2,3)

let r be the radius of circle

Plugging in this value of center in standard equation for circle given above we have

$(x-a)^2 + (y-b)^2 = r^2 \ substitute (a,b) \ with (-2,3) \\=>(x-(-2))^2 + (y-3)^2 = r^2 \\=>(x+2)^2 + (y-3)^2 = r^2 (1)$

Given that point (1,2 ) passes through circle. Hence this point will satisfy the above equation of circle.

Plugging in the point (1,2 ) in equation 1 we have

$\\=>(x+2)^2 + (y-3)^2 = r^2 \\=> (1+2)^2 + (2-3)^2 = r^2\\=> 3^2 + (-1)^2 = r^2\\=> 9 + 1 = r^2\\=> 10 = r^2\\=> r^2 = 10\\$

now we have value of r^2 = 10, substituting this in equation 1 we have

Thus complete equation of circle is $=>(x+2)^2 + (y-3)^2 = r^2\\=>(x+2)^2 + (y-3)^2 = 10$

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3 years ago