Answer:
P____Q____R
PR= PQ+ QR
(14x-13) = (5x-2)+(6x+1)
14x-13= 11x – 1
14x – 11x = 13–1
3x = 12
x= 12/ 3
x= 4
PR= 14x – 13 = 14 (4) – 13 = 18 – 13= 5
If you want (PQ , QR ) this is the solution
PQ =5x-2=5(4)-2=20-2=18
QR =6x+1=6(4)+1=24+1=25
I hope I helped you^_^
Answer: number of years that it will take for the balance to reach $120,000 is 42 years
Step-by-step explanation:
Initial amount deposited into the account is $4000. This means that the principal is $4000
P = 4000
It was compounded annually. This means that it was compounded once in a year. So
n = 1
The rate at which the principal was compounded is 8.4%. So
r = 8.4/100 = 0.084
Let the number of years that it will take for the balance to reach $120,000. It means that it was compounded for a total of t years.
Amount, A at the end of t years is $120,000
The formula for compound interest is
A = P(1+r/n)^nt
120000 = 4000(1 + 0.084/1)^1×t
120000/4000 = 1.084^t
30 = 1.084^t
t = 42 years
Answer:
There are 36 green marbles, 24 red marbles and 16 blue marbles.
Step-by-step explanation:
Answer:
4(x - 1) = 4x - 4
3x + 6 = 3(x + 2)
Step-by-step explanation:
The first equation is

We simplify to get;

This is not true, therefore this equation has no solution.
The second equation is

Combine like terms:



This has a unique solution.
The 3rd equation is

Group similar terms:

The 4th equation is :


This is always true. The equation has infinite solution.
The 5th equation is:

This also has infinite solution
The 6th equation is

It has a unique solution.
The equations which gave the same solution as; 2.3p – 10.1 = 6.5p – 4 – 0.01p are;
- 2.3p – 10.1 = 6.49p – 4 and
- 230p – 1010 = 650p – 400 – p
<h3>Which equations have the same solution as the given equation?</h3>
It follows from the answer choices that the equation; 2.3p – 10.1 = 6.49p – 4 is equivalent to the given equation as the algebraic sum of p variables is evaluated on the right hand side.
The second equation; 230p – 1010 = 650p – 400 – p also is equivalent to the given equation, when multiplied through by 100.
Read more on equivalent equations;
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