Answer: (7) 7.0 (8) 17.3 (9) 64.7°
<u>Step-by-step explanation:</u>
Law of Cosines is: a² = b² + c² - 2bc · cos A
or: b² = a² + c² - 2ac · cos B
or: c² = a² + b² - 2ab · cos C
7) b = 4, c = 10, ∠A = 34
a² = b² + c² - 2bc · cos A
= 4² + 10² - 2(4)(10) · cos 34°
= 16 + 100 - 80 (0.829)
= 116 - 66.323
= 49.677
a = 
a = 7.0
8) a = 11, b = 8, ∠C = 131
c² = a² + b² - 2ab · cos C
= 11² + 8² - 2(11)(8) · cos 131°
= 121 + 64 - 176 (-0.656)
= 185 + 115
= 300
c = 
c = 17.3
9) a = 10.5, b = 5.9, c = 9.6
9.6² = 10.5² + 5.9² - 2(10.5)(5.9) · cos C
92.16 = 110.25 + 34.81 - 2(10.5)(5.9) · cos C
92.16 = 145.06 - 123.9 cos C
-52.9 = -123.9 cos C
0.427 = cos C
cos⁻¹ (0.427) = C
64.7° = C
A. No x value can have multiple different outputs.
D. The change in y between x values is 4 but the function when x is 0 is -2
I believe the answer is D. $34.43
The orthocenter is the point where the three altitudes meet.
sketch the graph and you will see that AB is a horizontal line, the altitude is a vertical line through the point (1,3), so the equation of this altitude is x=1
next, find another altitude. I'll use the altitude of BC.
the slope of BC is (6-3)/(4-1)=1, so the slope of the altitude, which is perpendicular to BC going through the point A (0,6), is -1, the equation of the altitude of BC is y=-x+6
the system of equation : x=1
y=-x+6
has a solution (1, 5)
the solution is where the two lines meet, the meeting point is the orthocenter.
Double check by find the equation for other altitude:
slope of AC: (3-6)/(1-0)=-3
slope of altitude of AC: 1/3
equation of altitude of AC: y=(1/3)x+b
the altitude of AC goes through point B (4,6), so we can find out b by plug x=4, y=6 in the equation: 6=(1/3)*4+b, b=14/3
y=(1/3)x+14/3
Is (1,5) also a solution to this equation? Plug x=1 in the equation, we get y=5, so yes, (1,5) is a point on the third altitude.
Answer:
Given that,
The number of grams A of a certain radioactive substance present at time, in years
from the present, t is given by the formula
a) To find the initial amount of this substance
At t=0, we get
We know that e^0=1 ( anything to the power zero is 1)
we get,
The initial amount of the substance is 45 grams
b)To find thehalf-life of this substance
To find t when the substance becames half the amount.
A=45/2
Substitute this we get,
Taking natural logarithm on both sides we get,
Half-life of this substance is 154.02
c) To find the amount of substance will be present around in 2500 years
Put t=2500
we get,
The amount of substance will be present around in 2500 years is 0.000585 grams
