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1 answer:
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Answer:
The amount invested at 10% rate is $ 2200
Step-by-step explanation:
Given as :
The total amount earn by student = $6500
The total interest received at the end of year = $607
Let the amount invested at 10% rate = $x
So, The amount invested at 9% rate = ($6500 - $x)
Let the time for which money invested = 1 year
<u>Now from Simple Interest method :</u>
SI =
Or, SI_1 + SI_2 = +
Or, SI_1 + SI_2 =
Or, $607 × 100 = 10x + (6500-x) × 9
Or, $60700 = 10x- 9x + 58500
Or, $60700 - $58500 = x
Or , 2200 = x
∴ x = 2200
So, The amount invested at 10% rate = $x = $ 2200
Hence The amount invested at 10% rate is $ 2200 Answer
Answer:
(1/4)*(e⁶ - 7)
Step-by-step explanation:
a) Given
x − y = 0 if x = 0 ⇒ y = 0
x − y = 2 if x = 0 ⇒ y = -2; if y = 0 ⇒ x = 2
x + y = 0 if x = 0 ⇒ y = 0
x + y = 3 if x = 0 ⇒ y = 3; if y = 0 ⇒ x = 3
then we show the region R in the pics 1 and 2.
b) We make the change of variables as follows
u = x + y
v= x - y
If
x - y = 0 ⇒ v = 0
x − y = 2 ⇒ v = 2
x + y = 0 ⇒ u = 0
x + y = 3 ⇒ u = 3
Where u is the horizontal axis and v is the vertical axis, the new region S is shown in the pic 3.
c) We evaluate ∫∫R (x + y)*e∧(x² - y²)dA
The procedure is shown in the pic 4, where we have to calculate the Jacobian in order to use it to get the answer.
