Step-by-step explanation:
Let "c" be the original number of classrooms.
The 1200/c was the original number of students per classroom.
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Equation:
1200/(c-4) = (1200/c)+10
Multiply thru by c(c-4)
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1200c = 1200(c-4)+10c(c-4)
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1200c = 1200c-4800 + 10c^2-40c
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10c^2-40c-4800 = 0
c^2-4c-480 = 0
Factor:
(c-24)(c+20) = 0
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Positive solution:
c = 24 (# of classrooms originaly planned)
Answer:
At-most 3 birdhouses.
Step-by-step explanation:
Let B represent the number of birds and H represent the number of birdhouses that Nathaniel can build with his Lego blocks.
We have been given that Nathaniel builds birds and birdhouses using Lego blocks. The inequality
represents the number of birds and birdhouses Nathaniel could build.
To find the number of birdhouses that Nathaniel can build after making 50 birds, we will substitute B=50 in our given inequality and then solve for H.


Let us subtract 2150 from both sides of our inequality.

Let us divide both both sides of our inequality by 215.


As Nathaniel can make build less than or equal to 3.953488 and the biggest integer less than or equal to 3.953 is 3, therefore, Nathaniel can build at-most 3 birdhouses with remaining Lego blocks.
Answer:
First statement is correct
Step-by-step explanation:
Segment AB is tangent to circle C because
Im not quite sure but i think the unit rate is 150. Hope this helps!!