In triangle ABC, B = 90 degrees, and AC = 13 cm.
Determine:
A) BC if C = 17 degrees. B) A if BC = 6cm .
A) If C = 17° then A = 90 -17 = 73°
By the Law of Sines
BC / sin 73 = AC / sin90
BC / sin 73 = 13/1
BC = 13sin(73) ≈ 12.4
B) A if BC = 6
Again by the Law of Sines
sin A / BC = sin90 / AC
sin A / 6 = 1 /13
sin A = 6/13 take the sine inverse to find angle A
arcsin (6/13) = A ≈ 27.5°
Subtraction Property of Equality
Division Property of Equality
2.
For how many positive integers x is x^2+4x+4 between 10 and 50?
Note x^2 + 4x + 4 = (x + 2)^2
Whe x = 1 (x + 2)^2 = 9
When x = 2 (x + 2)^2 = 16
When x = 5 (x + 2)^2 = 49
When x = 6 (x + 2)^2 = 64
So.... four integers....2, 3 , 4 , 5
1.
Solve for x, where x>0 and 0=-9x^2-3x+2 Express your answer as a simplified common fraction.
Multiply through by - 1 and we have
9x^2 + 3x - 2 = 0
(3x + 2) (3x - 1) = 0
Only the second factor produces a positive solution of 1/3
Let me check!!!
The solutions are 1/2 + √5/2 and 1/2 - √5/2
So we have
[ (1/2 + √5/2) - (1/2 - √5/2) ]^2 =
[ √5/2 + √5/2 ]^2 =
[ 2√5 / 2 ]^2
[ √5 ] ^2 =
5
EDITED
3y^2 + 5y + 2 = 4
3y^2 + 5y - 2 = 0
(3y - 1) (y + 2) = 0
Set each factor to 0...solve for y
3y - 1 = 0 y + 2 = 0
3y =1 y = - 2
y =1/3
y = -2 is the least value
Ah, tertre....you beat me to it....sneaky !!!!
( x + y)^2 = 78
x^2 + 2xy + y^2 = 78
If xy = 13 .....then
x^2 + y^2 + 2(13) = 78
x^2 + y^2 + 26 = 78
x^2 + y^2 = 52
So
(x - y)^2 =
x^2 - 2xy + y^2 =
(x^2 + y^2) - 2xy =
(52) - 2(13) =
26
For what values of k does the equation 3x^2 + 4x + k = 0 have no real roots? One real root? Two real roots?
No real roots if the discriminant is < 0 ....so....
(-4)^2 - 4(3)(k) < 0
16 - 12k < 0
16 < 12k
4/3 < k
k > 4/3
One real root if the disriminant = 0
(-4)^2 - 4(3)k = 0
16 - 12k = 0
16 = 12k
4/3 = k
It's obvious that we must have two real roots if k < 4/3
