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andre [41]
3 years ago
10

Find the principle that yields and interest of Rs.2240 at the rate of 10% p.a. in a years 6 months.​

Mathematics
1 answer:
Vinvika [58]3 years ago
5 0

Step-by-step explanation:

I=PRT/100

p=I/RT×100

p=2240/10×2×100

p=11200

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A saxophone costs $1200. A store offers two loan options. Which options saves more money if you pay the loan in 2 years?
avanturin [10]
Option 2.)
(Please correct me if I'm wrong)
8 0
3 years ago
The quadrilateral shown is rotated 90° clockwise about the origin. In which quadrant is the image of the quadrilateral located?
Viktor [21]

Answer:

Option (2). 1

Step-by-step explanation:

Coordinates of point A, B, C and D are,

A(-4, 4), B(-2, 4), C(-2, 1) and D(-4, 3).

Quadrilateral ABCD when rotated 90° clockwise about the origin,

Rule for the rotation of the vertices,

(x, y) → (y, -x)

Following the rule of rotation coordinates of the image points,

A(-4, 4) → A'(4, 4)

B(-2, 4) → B'(4, 2)

C(-2, 1) → C'(1, 2)

D(-4, 3) → D'(3, 4)

Since all image points have the positive coordinates (x and y coordinates), image quadrilateral A'B'C'D' will be located in 1st quadrant.

Option (2) is the correct option.

4 0
2 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
2 years ago
Name the solid and find its volume. Use 3.14 as the approximate value of pi.
sineoko [7]

It is given in the question that the solid is a cone with height of 16 units and radius of 7 units .

The formula of volume of cone is

V = \frac{1}{3} \pi r^2 h

To find the volume, we substitute the given values of r and h and use 3.14 for pi. That is

V = \frac{1}{3} *3.14*7^2 *16 = 820.6 \ units \ cube

And that's the required volume of the cone .

7 0
3 years ago
Jennifer got a box of chocolates. The box is a right triangular prism shaped box. It is 7 inches long, and the triangular base m
Dafna1 [17]
The surface area would be 68 squared inches. I hope this helps!
4 0
3 years ago
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