—Some JHS MATH TUTORIALS—
1.) If 5^n + 5^n + 5^n + 5^n + 5^n = 5, what is n?
Since limang beses inadd, multiply by 5 diba?
5(5^n)=5
By the law of exponents, (1 ang exponent ni 5 diba?)
So magiging 5^(n+1)=5^(1)
Iequate mo si n+1 sa 1
n+1=1
n=0
2.) Given that f(3x+8) = x^2 + 5x - 3, find f(9)
Get the inverse function of y=3x+8.
y=3x+8
Interchange x and y then solve for y
x=3y+8
Subtract 3y to both sides.
x-3y=3y-3y+8
x-3y=8
Subtract x to both sides
x-x-3y=8-x
-3y=8-x
Divide both sides by -3
-3y/(-3)=(8-x)/(-3)
y=(8-x)/(-3)
y=(x-8)/3
Substitute 9 to x.
y=(9-8)/3
y=1/3
Substitute 1/3 to the original equation.
f(1/3)=(1/3)^2+5(1/3)-3
=1/9 + 5/3 - 3
=1/9 + 15/9 - 27/9
=(1+15-27)/9
=(16-27)/9
=-11/9
3.) The sum of the first five terms of an arithmetic sequence is 50 and the first term is 4. Find the 6th term.
Substitute the given to the sum of an arithmetic sequence.
S_n=n/2[2a_1+(n-1)d]
So meron kang S_5 (eto ung sum ng first 5 terms) at n=5 at a_1=4
50=5/2[2(4)+(5-1)d]
50=5/2(8+4d)
50=20+10d
10d+20=50
10d+20-20=50-20
10d=30
d=3
Substitute the given to
a_n=a_1+(n-1)d
Since 6th term hinahanap,
a_6=4+(6-1)3
a_6=4+5(3)
a_6=4+15
a_6=19
4.) Find the quadratic equation whose roots are [3+sqrt(3)]/7 and [3-sqrt(3)]/7
Let x=[3+sqrt(3)]/7 --> you can choose any of the two roots.
x=[3+sqrt(3)]/7
First thing to do is multiply both sides by 7 to eliminate the denominator.
7x=3+sqrt(3)
Subtrat 3 from both sides.
7x-3=3-3+sqrt(3)
7x-3=sqrt(3)
Square both sides
49x^2-42x+9=3
49x^2-42x+6=0
Dapat may equal to zero para masabing equation siya.
5.) The legs of a right triangle are 10 and 29. Find the secant of the smaller angle.
Draw ka ng right triangle. Always remember na ang side adjacent to the smaller angle is longer. (Go ahead, draw it)
By triangle trigonometry,
sec theta=(hyp)/(adj)
To find the hypotenuse, use the Pythagorean Theorem,
10^2+29^2=941
hyp=sqrt(941)
Thus, the secant of the smallest angle is
sec theta=[sqrt(941)]/29
