Ask question

Ask question

All categories

2 years ago

15

I borrowed 12000 rupees from jamshed at 6 percent per annum simple interest for 2 years .Had i borrwed this sum at 6 percent per

annum compound interest, what extra amount would i gave to pay ?

1 answer:

Leni [432]2 years ago

6 0

Answer:

43.2 Rupees

Step-by-step explanation:

To determine how much extra will be paid, we calculate the Amount at Simple Interest and the Amount at Compound Interest and Make Comparisons.

Now, Principal=12000

Rate=6%

Time= 2 years

Amount at Simple Interest= Principal+ Simple Interest

=P+(PRT)/100

=12000+(12000X6X2)/100

=12000+1440

=13440 Indian Rupees

Amount at Compound Interest

A=P(1+r)ⁿ

=12000(1+0.06)²

=12000(1.06)²

=13483.2 Indian Rupees

Difference=Amount at Compound Interest- Amount at Simple Interest

=13483.2-13440

=43.2 Indian Rupees

If he had borrowed at Compound interest, he would have paid an extra 43.2 Rupees

You might be interested in

The dot plot below shows how many chocolate chips were in each cookie Lauren brought to school.

Marta_Voda [28]

8+9+10= 27 and 27 divided by 3 is 9. So the mean of the chocolate chips is 9

5 0

1 year ago

Dominic bought a new alarm clock that was on sale for $18.75. if this price represents a 30% discount from the original price. w

fredd [130]

So here you are trying to find out what times .30 (30%) subtracted the original price will give you 18.75. Because it is 30% discount you want to use 70% (.7) so that you can find the price before not the price after.

so .7(x) = 18.75

Which you then divide and then get 26.79 dollars.

4 0

2 years ago

Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One common ch

enot [183]

Answer:

$y=\frac{-7t^2+22t-7}{7t-22}$

Step-by-step explanation:

We are given that

Initial value problem

$y'=(t+y)^2-1$ , y(3)=4

Substitute the value $z=t+y$

When t=3 and y=4 then

z=3+4=7

$y'=z^2-1$

Differentiate z w.r.t t

Then, we get

$\frac{dz}{dt}=1+y'$

$z'=1+z^2-1=z^2$

$z^{-2}dz=dt$

Integrate on both sides

$-\frac{1}{z}dz=t+C$

$z=-\frac{1}{t+C}$

Substitute t=3 and z=7

Then, we get

$7=-\frac{1}{3+C}$

$21+7C=-1$

$7C=-1-21=-22$

$C=-\frac{22}{7}$

Substitute the value of C then we get

$z=-\frac{1}{t-\frac{22}{7}}$

$z=\frac{-7}{7t-22}$

$y=z-t$

$y=\frac{-7}{7t-22}-t$

$y=\frac{-7-7t^2+22t}{7t-22}$

$y=\frac{-7t^2+22t-7}{7t-22}$

8 0

2 years ago

ANSWER ALL 5 PARTS.

N76 [4]

A function $f$ from a set A to a set B is defined as a relation that assings to each element $x$ in the set A exactly one element $y$ in the set B. The set A is called the domain of the function while the set B is the range. So we have five statements and need to find some functions. Melissa decides to reserve a patch in her vegetable garden for growing bell peppers. If each side of the tomato patch is $x$ feet, then we have a square patch as shown in the Figure below.

1.a) Write the function Wa(x) representing the width of the bell pepper patch.

We know that she wants its width to be half the width of the tomato patch. Let $x$ be the width of the tomato patch, then the function that matches this statement is:

$\boxed{Wa(x)=\frac{x}{2}}$

1.b) Write the function La(x) representing the length of the bell pepper patch.

In this case Melissa wants <span>its length to exceed the length of the tomato patch by 2 feet. To do this we enlarge the length of the tomato patch 2 feet. Therefore the function is the following:

</span> $\boxed{La(x)=x+2}$
<span>
2. Ar</span>ea of the bell pepper patch in terms of x.

Given that the bell pepper patch is a rectangle, then t<span>he area of a rectangle is the product of the length and width. So:

</span> $A=(\frac{x}{2})(x+2) \\ \\ \therefore \boxed{A=\frac{x(x+2)}{2}}$
<span>
3. C</span><span>ombined area of the tomato patch and the bell pepper patch.

This function is the sum of both the area of the tomato patch and the bell pepper patch. So:

</span> $Aar(x)=x^{2}+\frac{x(x+2)}{2} \rightarrow Aar(x)=x^{2}+ \frac{x^{2}}{2}+x \rightarrow Aar(x)=\frac{3x^{2}}{2}+x \\ \\ \therefore \boxed{Aar(x)=\frac{x(3x+2)}{2}}$
<span>
4. W</span>rite the function Aa(x) for the remaining planting area in the garden.

The remaining planting area in the garden are the rectangles in red. So we need to subtract the width of the bell pepper patch from the width of the tomato patch and multiply it by 2. In mathematical language this is given by:<span>

</span> $Aa(x)=2(x-\frac{x}{2}) \rightarrow Aa(x)=x$

5. Find the area of the remaining space in the garden after planting tomatoes and bell peppers.

Given that <span>Melissa wants the area of the bell pepper patch to be 31.5 square feet, then it is true that:

</span> $31.5=\frac{x(x+2)}{2} \rightarrow x^{2}+2x-64=0 \\ \\ solving \ for \ x: \\ x=7.06$
<span>
Therefore the area of the remaining space is:

</span> $\boxed{Aa(7.06)=7.06ft^{2}}$

6 0

2 years ago

a barangay has 1000 individuals and its population doubles every 60 years. Give an exponential model for the barangay's populati

sdas [7]

Answer:

P(t) = 1000e^(0.01155)t

Step-by-step explanation:

Let the population of barangay be expressed according to the exponential formula;

P(t) = P0e^kt

P(t) is the population of the country after t years

P0 is the initial population

t is the time

If barangay has 1000 initially, this means that P0 = 1000

If the population doubles after 60years then;

at t = 60, P(t) = 2P0

Substitute into the formula

2P0 = P0e^k(60)

2 = e^60k

Apply ln to both sides

ln2 = lne^60k

ln2 = 60k

k = ln2/60

k = 0.01155

Substitute k = 0.01155 and P0 into the expression

P(t) = 1000e^(0.01155)t

Hence an exponential model for barangay's population is

P(t) = 1000e^(0.01155)t

7 0

2 years ago