-20 because 40 is a positive and spent is negative
Answer:
1.)
≈ 3.652
2.) I would say something about how the A in front of cos in the equation would change to 90, rather than stay 75 (in the equation for the step by step), but it would be easier to just use the Pythagorean theorem.
Step-by-step explanation:
I think we may have the same class so hopefully this helps:
1.)
--> law of cosines formula.
--> plugged in numbers; when you draw the triangle, the included angle would be A, and the opposite side would be a. B and b, and C and c are opposite each other. In this case, a is the hypotenuse.
--> in between steps.
--> more simplifying.
--> answer
2.) This one is just an explanation: The 75 in the equation is the given angle, which is a. If this changes, it would just change in the equation too. And obviously, if it's 90 degrees, you can just use Pythagorean theorem a^2+b^2=c^2.
Good luck! :)
Answer:
The average annual growth rate of a certain country's population for 1950, 1988, and 2010 are 2.398, 0.9985 and 0.2236 respectively.
Step-by-step explanation:
The given equation is

Where Y is the annual growth rate of a certain country's population and x is the number of years after 1900.
Difference between 1950 and 1900 is 50.
Put x=50 in the given equation.


Therefore the estimated average annual growth rate of the country's population for 1950 is 2.398.
Difference between 1988 and 1900 is 88.
Put x=88 in the given equation.


Therefore the estimated average annual growth rate of the country's population for 1988 is 0.9985.
Difference between 2010 and 1900 is 110.
Put x=110 in the given equation.


Therefore the estimated average annual growth rate of the country's population for 2010 is 0.2236.
Answer:
603 miles in 3 days
Step-by-step explanation:
Let's find out how many miles in 1 day:
- 1,005 ÷ 5 = 201
- So, Sarah drives 201 miles in 1 day
Let's find out how many miles in 3 days:
- 3 days = 1 day × 3
- 3 days = 201 × 3
- 3 days = 603 miles
I hope this helps!
Answer:
A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S.
The given equation consisting of variable , m is
![\frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})](https://tex.z-dn.net/?f=%5Cfrac%7B2%20m%7D%7B2%20m%2B3%7D%20-%5Cfrac%7B2%20m%7D%7B2%20m-3%7D%3D1%5C%5C%5C%5C%202%20m%5B%5Cfrac%7B1%7D%7B2%20m%2B3%7D%20-%5Cfrac%7B1%7D%7B2%20m-3%7D%5D%3D1%5C%5C%5C%5C%202%20m%5Ctimes%20%5Cfrac%7B%5B2%20m-3%20-2%20m-%203%5D%7D%7B4m%5E2-9%7D%3D1%5C%5C%5C%5C%20-6%20%5Ctimes%202%20m%3D4%20m%5E2%20-9%5C%5C%5C%5C%204%20m%5E2%20%2B1%202%20m%20-9%3D0%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%5Csqrt%7B12%5E2-4%20%5Ctimes%204%20%5Ctimes%20%28-9%29%7D%7D%7B2%5Ctimes%204%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%20%5Csqrt%20%7B144%2B144%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%20%5Csqrt%20%7B288%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%2012%20%5Csqrt%7B2%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B3%7D%7B2%7D%5Ctimes%28-1%20%5Cpm%20%5Csqrt%7B2%7D%29)
None of the two solution
, is extraneous.
Here, L.H.S= R.H.S
Option A: 0→ extraneous