Answer:
7 is in the hundred thousandth place
2 is in the hundredth place
1 is in the ten thousandth place
Answer:
Both a = −4 and a = −1 are true solutions
Step-by-step explanation:
Given
![\sqrt{a + 5} = a + 3](https://tex.z-dn.net/?f=%5Csqrt%7Ba%20%2B%205%7D%20%3D%20a%20%2B%203)
![a = -4; a = -1](https://tex.z-dn.net/?f=a%20%3D%20-4%3B%20a%20%3D%20-1)
Required
The true statement about the solutions
We have:
![\sqrt{a + 5} = a + 3](https://tex.z-dn.net/?f=%5Csqrt%7Ba%20%2B%205%7D%20%3D%20a%20%2B%203)
Square both sides
![a + 5 = (a + 3)^2](https://tex.z-dn.net/?f=a%20%2B%205%20%3D%20%28a%20%2B%203%29%5E2)
![a + 5 =a^2 + 6a + 9](https://tex.z-dn.net/?f=a%20%2B%205%20%3Da%5E2%20%2B%206a%20%2B%209)
Collect like terms
![a^2 + 6a - a + 9 - 5 = 0](https://tex.z-dn.net/?f=a%5E2%20%2B%206a%20-%20a%20%2B%209%20-%205%20%3D%200)
![a^2 + 5a + 4 = 0](https://tex.z-dn.net/?f=a%5E2%20%2B%205a%20%2B%204%20%3D%200)
Expand
![a^2 + 4a + a + 4 = 0](https://tex.z-dn.net/?f=a%5E2%20%2B%204a%20%2B%20a%20%2B%204%20%3D%200)
Factorize
![a(a + 4) + 1(a + 4) = 0](https://tex.z-dn.net/?f=a%28a%20%2B%204%29%20%2B%201%28a%20%2B%204%29%20%3D%200)
Factor out a + 4
![(a + 1)(a + 4) = 0](https://tex.z-dn.net/?f=%28a%20%2B%201%29%28a%20%2B%204%29%20%3D%200)
Split:
![a + 1 = 0; a + 4 = 0](https://tex.z-dn.net/?f=a%20%2B%201%20%3D%200%3B%20a%20%2B%204%20%3D%200)
Solve:
![a =-1; a = -4](https://tex.z-dn.net/?f=a%20%3D-1%3B%20a%20%3D%20-4)
Ah...Trigonometry is fun!
The law of sines states:
![\frac{sinA}{a}= \frac{sinB}{b} = \frac{sinC}{c}](https://tex.z-dn.net/?f=%20%5Cfrac%7BsinA%7D%7Ba%7D%3D%20%5Cfrac%7BsinB%7D%7Bb%7D%20%3D%20%5Cfrac%7BsinC%7D%7Bc%7D%20)
The transitive property (switching the orders of the equations) applies here. Therfore, we can say that
![\frac{sinA}{a}=\frac{sinC}{c}](https://tex.z-dn.net/?f=%5Cfrac%7BsinA%7D%7Ba%7D%3D%5Cfrac%7BsinC%7D%7Bc%7D%20)
We then plug in our given values to find C
![\frac{sin24}{77}=\frac{sinC}{162}](https://tex.z-dn.net/?f=%5Cfrac%7Bsin24%7D%7B77%7D%3D%5Cfrac%7BsinC%7D%7B162%7D%20)
![\frac{162*sin24}{77}=sinC](https://tex.z-dn.net/?f=%5Cfrac%7B162%2Asin24%7D%7B77%7D%3DsinC)
Solving, we get 0.8557316387.
We're not done yet!We are trying to find an angle measure, so we'll do the inverse of the ratio we used (sin).
arcsin0.8557316387 (arcsin is the same as inverse sin)
=
58.8 (approximate)
So the measure of angle C is 58.8. You could check this by reinserting it into the equation
![\frac{sinA}{a}= \frac{sinB}{b} = \frac{sinC}{c}](https://tex.z-dn.net/?f=%20%5Cfrac%7BsinA%7D%7Ba%7D%3D%20%5Cfrac%7BsinB%7D%7Bb%7D%20%3D%20%5Cfrac%7BsinC%7D%7Bc%7D%20)
.
:)
The answer is option c.
That is, the wrong step in step 6. It was written that the center of the circunference is the point (2.1). However, the general equation of a circumference is:
(X- (a)) ^ 2 + (Y- (b)) ^ 2 = r ^ 2
Where the point (a, b) is the center of the circle.
So for this case the point for the center is: (-2, -1)