SOLUTION: I need help for each of the algebra problems below.
1. 200= 210/(1+x)
2. 110.00 = 133.10/(1+x)2
3. 0.08243=〖[1+x/4]〗4 - 1
4. 34.99=100ex
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-> SOLUTION: I need help for each of the algebra problems below.
1. 200= 210/(1+x)
2. 110.00 = 133.10/(1+x)2
3. 0.08243=〖[1+x/4]〗4 - 1
4. 34.99=100ex
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Question 263566: I need help for each of the algebra problems below.
1. 200= 210/(1+x)
2. 110.00 = 133.10/(1+x)2
3. 0.08243=〖[1+x/4]〗4 - 1
4. 34.99=100ex Found 2 solutions by drk, JBarnum:Answer by drk(1908) (Show Source):
You can put this solution on YOUR website! 1. 200= 210/(1+x)
step 1 - multiply by 1+x to get
distribute to get
subtract to get
divide to get
-----
2. 110.00 = 133.10/(1+x)^2
step 1 - multiply by (1+x)^2 to get
step 2- divide by 110 to get
square root both sides to get
subtract 1 to get
so,
x = -.1
or
x = -2.1
-----
3. 0.08243=〖[1+x/4]〗4 - 1
--> not sure what we have on this one :((
----
4. 34.99=100e^x
step 1 - divide by 100 to get
step 2 - ln both sides to get
You can put this solution on YOUR website! 1. 200= 210/(1+x)
2. 110.00 = 133.10/(1+x)2
3. 0.08243=〖[1+x/4]〗4 - 1
4. 34.99=100ex
1. 200=210/(1+x) Get x to the numerator by multiplying each side by the denomerator
(1+x)200=210/(1+x)(1+x) {ex 2/3(3)...(2/3)(3/1)=2}
(1+x)200=210 Multiply
200+200x=210 Subtract 200
-200 -200
200x=10 Divide by 200
x=10/200 Simplify
x=1/20
Answer check:
200=210/(1+1/20) 1/20=.05
200=210/(1+.05) Add
200=210/1.05 Divide
200=200
2. 110=133.10/(1+x)2
this is just like #1 just multiply the 2 first then the denominator
(1+x)2=(2+2x)
110=133.10/(2+2x)
(2+2x)110=133.10/(2+2x)(2+2x)
(2+2x)110=133.10
220+220x=133.10 Subtract 220
-220 -220
220x=-86.9 Divide by 220
x=-86.9/220
x= -.395
Check answer:
110=133.10/(1-.395)2 Subtract
110=133.10/(.605)2 Multiply
110=133.10/(1.21) Divide
110=110
3. unknown symbols used cant answer it
4. 34.99=100ex
Find for e then plug in and solve for x to get e
34.99=100ex
34.99/x=100ex/100x
34.99/100x=e
34.99/100e=x
is the question looking for a number for x and e or as I have left them. there needs to be another equation for me to solve for e or x as a single number.