What is the answer to this problem?
1 answer:
Answer:
<h2>48</h2>Step-by-step explanation:
It's an arithmetic sequence with common difference <em>d = 7</em>.
Find 9th term.
METHOD 1<em>.</em>
<em>Calculate next terms:</em>
METHOD 2.
<em>The explicit formula of an arithmetic sequence:</em>
Substitue:
<em>use the distributive property</em>
Put <em>n = 9</em>:
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To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]