Please help, I’m so close to passing math
1 answer:
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Answer:
1) (x + 3)(3x + 2)
2) x= +/-root6 - 1 by 5
Step-by-step explanation:
3x^2 + 11x + 6 = 0 (mid-term break)
using mid-term break
3x^2 + 9x + 2x + 6 = 0
factor out 3x from first pair and +2 from the second pair
3x(x + 3) + 2(x + 3)
factor out x+3
(x + 3)(3x + 2)
5x^2 + 2x = 1 (completing squares)
rearrange the equation
5x^2 + 2x - 1 = 0
divide both sides by 5 to cancel out the 5 of first term
5x^2/5 + 2x/5 - 1/5 = 0/5
x^2 + 2x/5 - 1/5 = 0
rearranging the equation to gain a+b=c form
x^2 + 2x/5 = 1/5
adding (1/5)^2 on both sides
x^2 + 2x/5 + (1/5)^2 = 1/5 + (1/5)^2
(x + 1/5)^2 = 1/5 + 1/25
(x + 1/5)^2 = 5 + 1 by 25
(x + 1/5)^2 = 6/25
taking square root on both sides
root(x + 1/5)^2 = +/- root(6/25)
x + 1/5 = +/- root6 /5
shifting 1/5 on the other side
x = +/- root6 /5 - 1/5
x = +/- root6 - 1 by 5
x = + root6 - 1 by 5 or x= - root6 - 1 by 5
Answers:
A = 120
b = 45.0
c = 33.2
Side Note: only one triangle is possible
See attached for a visual. I used GeoGebra to draw the triangle.
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Explanation:
We are given the following information
B = 35
C = 25
a = 68
We need to find the following
A, b, c
where the lower case letters represent the side lengths; the upper case letters are the angles. The angles are opposite their corresponding sides. For instance, side lowercase b is opposite angle uppercase B. The other letters are positioned the same way.
-----------------------
First use the idea that for any triangle, the three angles (A,B,C) must add to 180 degrees
A+B+C = 180
Replace B and C with 35 and 25. Solve for angle A
A+35+25 = 180
A+60 = 180
A+60-60 = 180-60
A = 120
Now that we know the three angles A = 120, B = 35, C = 25, we can find the missing sides 'b' and 'c'
-----------------------
We will use the law of sines to find side b
sin(A)/a = sin(B)/b
sin(120)/68 = sin(35)/b
b*sin(120) = 68*sin(35) <<--- cross multiply
b = 68*sin(35)/sin(120)
b = 45.037013350222 <<--- use a calculator; this value is approximate
b = 45.0 <<--- round to the nearest tenth
-----------------------
Do the same for side c
sin(A)/a = sin(C)/c
sin(120)/68 = sin(25)/c
c*sin(120) = 68*sin(25) <<--- cross multiply
c = 68*sin(25)/sin(120)
c = 33.1838323365404 <<--- use a calculator; this value is approximate
c = 33.2 <<--- round to the nearest tenth
A = 120
b = 45.0
c = 33.2
Side Note: only one triangle is possible
See attached for a visual. I used GeoGebra to draw the triangle.
-------------------------------------------------------------
-------------------------------------------------------------
Explanation:
We are given the following information
B = 35
C = 25
a = 68
We need to find the following
A, b, c
where the lower case letters represent the side lengths; the upper case letters are the angles. The angles are opposite their corresponding sides. For instance, side lowercase b is opposite angle uppercase B. The other letters are positioned the same way.
-----------------------
First use the idea that for any triangle, the three angles (A,B,C) must add to 180 degrees
A+B+C = 180
Replace B and C with 35 and 25. Solve for angle A
A+35+25 = 180
A+60 = 180
A+60-60 = 180-60
A = 120
Now that we know the three angles A = 120, B = 35, C = 25, we can find the missing sides 'b' and 'c'
-----------------------
We will use the law of sines to find side b
sin(A)/a = sin(B)/b
sin(120)/68 = sin(35)/b
b*sin(120) = 68*sin(35) <<--- cross multiply
b = 68*sin(35)/sin(120)
b = 45.037013350222 <<--- use a calculator; this value is approximate
b = 45.0 <<--- round to the nearest tenth
-----------------------
Do the same for side c
sin(A)/a = sin(C)/c
sin(120)/68 = sin(25)/c
c*sin(120) = 68*sin(25) <<--- cross multiply
c = 68*sin(25)/sin(120)
c = 33.1838323365404 <<--- use a calculator; this value is approximate
c = 33.2 <<--- round to the nearest tenth
