Given:
The two functions are:
To find:
The value of 
.
Solution:
We have,
We know that,
![(h\circ g)(b)=[(5b-9)-1]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28b%29%3D%5B%285b-9%29-1%5D%5E2)
![(h\circ g)(b)=[5b-10]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28b%29%3D%5B5b-10%5D%5E2)
Putting 
, we get
![(h\circ g)(-6)=[5(-6)-10]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28-6%29%3D%5B5%28-6%29-10%5D%5E2)
![(h\circ g)(-6)=[-39-10]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28-6%29%3D%5B-39-10%5D%5E2)
![(h\circ g)(-6)=[-49]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28-6%29%3D%5B-49%5D%5E2)
Therefore, the value of is 2401.
To solve a triangle means to obtain al the missing parts of the triangle.
<h3>How do you solve a triangle?</h3>
To solve a triangle means to obtain al the missing parts of the triangle. Now we have the following information;
<ABC = 51°
c = 8
a = 11
Then;
b^2 = a^2 + c^2 - 2acCos B
b^2 = (11)^2 + (8)^2 - [2 * 11 * 8 * cos 51]
b^2 = 121 + 64 - 55.38
b = 11.39
Now;
a/sinA = b/sinB
asinB = bsinA
sinA = asinB/b
sinA = 11 * sin 51/11.39
A = 48.6 or 49°
C = 180 - (49 + 51)
C = 80°
Learn more about triangles:brainly.com/question/2773823
#SPJ1
Answer:
Step-by-step explanation:
A1. C = 104°, b = 16, c = 25
Law of Sines: B = arcsin[b·sinC/c} ≅ 38.4°
A = 180-C-B = 37.6°
Law of Sines: a = c·sinA/sinC ≅ 15.7
A2. B = 56°, b = 17, c = 14
Law of Sines: C = arcsin[c·sinB/b] ≅43.1°
A = 180-B-C = 80.9°
Law of Sines: a = b·sinA/sinB ≅ 20.2
B1. B = 116°, a = 11, c = 15
Law of Cosines: b = √(a² + c² - 2ac·cosB) = 22.2
A = arccos{(b²+c²-a²)/(2bc) ≅26.5°
C = 180-A-B = 37.5°
B2. a=18, b=29, c=30
Law of Cosines: A = arccos{(b²+c²-a²)/(2bc) ≅ 35.5°
Law of Cosines: B = arccos[(a²+c²-b²)/(2ac) = 69.2°
C = 180-A-B = 75.3°
Answer:
$240
Step-by-step explanation:
I = prt
I = (2,400) (0.04) (2.5)
note: I changed 4% to a decimal and 30 months to 2.5 years
I = 96 (2.5)
I = 240 <-- The interest
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The simple interest accumulated on a principal of $ 2,400.00 at a rate of 4% per year for 2.5 years (30 months) is $240.00.
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* Last time I did interest was in 6th grade and I don't remember much. So I am very sorry if my answer is wrong *
