If the teacher has 30 students and 10% left,how much students does she have now
2 answers:
Answer:
She has 27 students now.
Step-by-step explanation:
Given :
The teacher has 30 students
10% students left.
To find : how much students does she have now?
Solution :
Since we are given that teacher has 30 students .
And 10% of students left
⇒ 10% of 30
⇒
⇒
⇒
Thus, 3 students left .
She had 30 students . three students left.
So, now she have 30 - 3 = 27 students .
Hence she has 27 students now.
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We know that
The arrangement forms an isosceles triangle with equal legs of 8 miles.
The angle between the legs is equal to
°
Therefore, the other two angles are
Angles = (180-60)/2 = 120/2 = 60°
It can, therefore, be noted that all angles are equal and thus the resulting triangle is actually an equilateral triangle and thus all the sides are equal.
Hence
the answer is
the distance between the two ships is 8 miles apart
alternative Method
Applying the law of cosines
<span>c²=a²+b²-2*a*b*cos C
</span>where
a=8 miles
b=8 miles
C is the angle between the legs-------> 123-63------> 60 degrees
c is the distance between the two ships
so
c²=8²+8²-2*8*8*cos 60------> c²=64-------> c=√64------> c=8 miles
The arrangement forms an isosceles triangle with equal legs of 8 miles.
The angle between the legs is equal to
Therefore, the other two angles are
Angles = (180-60)/2 = 120/2 = 60°
It can, therefore, be noted that all angles are equal and thus the resulting triangle is actually an equilateral triangle and thus all the sides are equal.
Hence
the answer is
the distance between the two ships is 8 miles apart
alternative Method
Applying the law of cosines
<span>c²=a²+b²-2*a*b*cos C
</span>where
a=8 miles
b=8 miles
C is the angle between the legs-------> 123-63------> 60 degrees
c is the distance between the two ships
so
c²=8²+8²-2*8*8*cos 60------> c²=64-------> c=√64------> c=8 miles
Answer:
Option C. -3x^2-13x-5
Step-by-step explanation:
we know that
To find how much greater is the area of room A than the area of room B. subtract the area of room B from the area of room A
so