Answer:
10. 9
In the first question we can solve using the Pythagorean theorem. It states that the hypotenuse in a right triangle or the longest side of the triangle, squared is equal to the other 2 sides, squared. Its expressed as so: C^2 = A^2 + b^2 where c is the hypotenuse and a and b are the other sides oft eh triangle.
Therefore,
15^2 = 12^2+x^2
225=144+x^2
81=x^2
x=9
__________________________________________________________
11. x=12
In this figure, there are 2 right triangles on the sides of the square. If we can find the lengths of the base of both triangles then we can find x using the Pythagorean theorem. the total base is 21 and the square is 11. 11+a+b=21. We can assume that both triangles are congruent and therefore we can solve this equation:
11+a+b=21
a+b=10
5+5=10
b=5
a=5
The bases of the two triangles are 5 and the hypotenuse is 13. Now we can solve using the Pythagorean theorem:
c^2 = a^2+b^2
13^2=5^2+x^2
169=25+x^2
144=x^2
144= 12 x 12
x=12
__________________________________________________________
12.
The diagnol of a triangle is the hypotenuse and since the length is the square root of 3, we have all the information we need.
c^2 = a^2+b^2
2^2 = (square root of 3)^2+b^2
4=3+b^2
1=b^2
b=1
Step-by-step explanation:
I think it has to be 2,333 students per year.
14,000 students divided by 6 years equals about 2,333 or 2,333.333.
Each successive number is divided by 3.
and so on. The next number in the sequence is
, and in general, you have a geometric sequence defined recursively by
and a closed form
