I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have
C(8,3) = 8!/(8-3)!3!
C(8,3) = 56
There are 56 3-person teams that can be formed from the 8 students.
Answer:
NQ = 26 cm
Step-by-step explanation:
A line bisector divides a line into two equal segments.
Given that line l is the bisector of segment NQ, it then means it divides NQ into two equal segments, namely, segment NP and segment PQ.
Thus, NP = ½ of segment NQ
Therefore, if NP = 13 cm,
13 = ½*NQ
multiply both sides by 2
2*13 = NQ
26 = NQ
NQ = 26 cm
Answer:
(x+1)(2x+5)
Step-by-step explanation:
f(x) = 2x² + 7x + 5
Factor the expression by grouping. First, the expression needs to be rewritten as 2x²+ax+bx+5. To find a and b, set up a system to be solved.
a+b=7
ab=2×5=10
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.
1,10
2,5
Calculate the sum for each pair.
1+10=11
2+5=7
The solution is the pair that gives sum 7.
a=2
b=5
2x²+7x+5 as (2x²+2x)+(5x+5).
(2x²+2x)+(5x+5)
Factor out 2x in the first and 5 in the second group.
2x(x+1)+5(x+1)
Factor out common term x+1 by using distributive property.
(x+1)(2x+5)
Answer:
1,-3,-5
Step-by-step explanation:
Given:
f(x)=x^3+7x^2+7x-15
Finding all the possible rational zeros of f(x)
p= ±1,±3,±5,±15 (factors of coefficient of last term)
q=±1(factors of coefficient of leading term)
p/q=±1,±3,±5,±15
Now finding the rational zeros using rational root theorem
f(p/q)
f(1)=1+7+7-15
=0
f(-1)= -1 +7-7-15
= -16
f(3)=27+7(9)+21-15
=96
f(-3)= (-3)^3+7(-3)^2+7(-3)-15
= 0
f(5)=5^3+7(5)^2+7(5)-15
=320
f(-5)=(-5)^3+7(-5)^2+7(-5)-15
=0
f(15)=(15)^3+7(15)^2+7(15)-15
=5040
f(-15)=(-15)^3+7(-15)^2+7(-15)-15
=-1920
Hence the rational roots are 1,-3,-5 !
