32 students.
3 do none, so we can leave them out.
29 students.
18 + 16 = 34
34 - 29 = 5 students doing both subjects.
5

+10-15=30
5

=35
<span>

=7
</span>х=49
5√49+10-15=30
30=30

+1+7=4
<span>

=-4
</span>it is <span>no solution
</span>
Answer: Hello, it would be 3(x-2)^2-(x-5)^2
Step-by-step explanation:
Answer:
a_n=125×20^(n-1)
Step-by-step explanation:
a_n=x×y^(n-1)
a_3=50000=x×y^2
a_5=20000000=x×y^4
y^2=20000000/50000=400
y=20;x=125
Answer:
No, it is not a square
Step-by-step explanation:
If one wall is 19", that would mean the wall perpendicular to this wall is also 19" (in fact all of the walls would be 19"!) If this was a square, then the diagonal we draw at 20.62" would serve as the hypotenuse of a right triangle. One wall would serve as a leg, and another wall as another leg. If this is a square, then the Pythagorean's Theorem would be satisfied when we plug in the 2 wall measures for a and b, and the diagonal for c:

We need to see if this is a true statement. If the left side equals the right side, then the 2 legs of the right triangle are the same length, and the room, then is a square.
361 + 361 = 425.1844
Is this true? Does 722 = 425.1844? Definitely not. That means that the room is not a square.