Hat is the equation of the quadratic function with a vertex at (2,–25) and an x-intercept at (7,0)?
1 answer:
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Answer:
Step-by-step explanation:
The relevant rule of exponents is ...
(a^b·c^d)^e = a^(be)·c^(de)
Then ...
(m^(5/4)·n^(-4/5))^(7/3) = m^(5/4·7/3)·n^(-4/5·7/3)
= m^(35/12)·n^(-28/15)
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Since you want positive rational exponents, you can write this as ...
= m^(35/12)/n^(28/15)
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'
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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