Answer:
The formula is
A=p (1+r/k)^kt
A future value?
P present value 4100
R interest rate 0.04
K compounded monthly 12
T time 10 years
A=4,100×(1+0.04÷12)^(12×10)
A=6,112.41. ..answer
Step-by-step explanation:
Answer:
d - 7
Step-by-step explanation:
The word "difference" tells us that we will be using subtraction. Also, subtraction is not the same if you reverse the terms like, x - 2 and 2 - x are not the same. In your question it say "d and 7" so that's the order we use in the algebraic expression.
Answer:
2x3 - 3x + 4
———————
x2
Step-by-step explanation:
Step 1 :
2
Simplify ——
x2
Rewriting the whole as an Equivalent Fraction :
4.1 Adding a fraction to a whole
Rewrite the whole as a fraction using x2 as the denominator :
x x • x2
x = — = ——————
1 x2
Let us test ....
P Q P/Q F(P/Q) Divisor
-1 1 -1.00 5.00
-1 2 -0.50 5.25
-2 1 -2.00 -6.00
-4 1 -4.00 -112.00
1 1 1.00 3.00
1 2 0.50 2.75
2 1 2.00 14.00
4 1 4.00 120.00
Polynomial Roots Calculator found no rational roots
Final result :
2x3 - 3x + 4
————————————
x2
Processing ends successfully
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Answer:
0.5
Step-by-step explanation:
Ok, so it's asking for what (1/(x-1) - 2/(x^2-1)) approaches as x approaches 1. Before we deal with the limit, let's simplify the inside.
We want to combine the two fractions into one fraction. Therefore, we need a common denominator.
1/(x-1) is equal to (x+1)/((x+1)(x-1) is equal to (x+1)/(x^2-1).
the inside expression is therefore (x+1)/(x^2-1) - 2/(x^2-1)
which simplifies to (x-1)/(x^2-1).
and that simplifies further to 1/(x+1).
Now this is a continuous function when x = 1, so to find the limit as x approaches 1 of this function, we can by definition just plug 1 in.
limx->1 (1/(x+1)) = 1/2.
The reason why we didn't just plug 1 in at the beginning is because the function wasn't continuous when x was 1.
Answer:
B/2
Step-by-step explanation:
The table mentions that the time actually starts from the end of the first month, so it cannot be graph one, which starts from 0.
Also, we can see that the values are decreasing over time, meaning it cannot be graph 3, which shows an increase.
thus, we settle the Answer as Graph 2, which satisfies all the needs.
Hope this helps.
