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3 years ago

How much interest is earned on a pricipal of $267 invested at an interest rate of %6 for six years?

Mathematics

1 answer:

Marianna [84]3 years ago

5 0

The correct answer is, 16.02.

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Ples help me find slant assemtotes

FrozenT [24]

A polynomial asymptote is a function $p(x)$ such that

$\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0$

$(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}$

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form $p(x)=ax+b$ .

Ignore the negative root (we don't need it). If $y=2x-1+2\sqrt{x^2-x}$ , then we want to find constants $a,b$ such that

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

We have

$\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}$

since $x\to\infty$ forces us to have $x>0$ . And as $x\to\infty$ , the $\dfrac1x$ term is "negligible", so really $\sqrt{x^2-x}\approx x$ . We can then treat the limand like

$2x-1+2x-ax-b=(4-a)x-(b+1)$

which tells us that we would choose $a=4$ . You might be tempted to think $b=-1$ , but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of $b$ , we have to solve for it in the following limit.

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0$

$\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2$

We write

$(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}$

Now as $x\to\infty$ , we see this expression approaching $-\dfrac12$ , so that

$-\dfrac12=\dfrac{b+1}2\implies b=-2$

So one asymptote of the hyperbola is the line $y=4x-2$ .

The other asymptote is obtained similarly by examining the limit as $x\to-\infty$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

$\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0$

Reduce the "negligible" term to get

$\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0$

Now we take $a=0$ , and again we're careful to not pick $b=-1$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0$

$\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2$

$(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
 x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}$

This time the limit is $\dfrac12$ , so

$\dfrac12=\dfrac{b+1}2\implies b=0$

which means the other asymptote is the line $y=0$ .

4 0

3 years ago

Question 10

KonstantinChe [14]

Answer:

area of the trapezoid=1/2(7+12)×5

=47.5cm

7 0

3 years ago

On Monday, Amanda runs 1 1/6 miles, and on Tuesday, she runs 2 1/2miles. How far did she run on Monday and Tuesday combined?

prohojiy [21]

Answer:

3 2/3 miles

Step-by-step explanation:

6 0

2 years ago

Ratio and simplify the ratio helpp

drek231 [11]

1: -4/1
2: 1/2
3: 1/4
4: -4/1

I hope those r the correct answers

8 0

3 years ago

Zip codes for the Houston area are five digit numbers with 7 as the first 2 digits of each zip code. The third digit can be 0, 3

djyliett [7]

Answer:

There are 400 possible zip codes in the Houston area

Step-by-step explanation:

Here, we want to calculate the possible number of zip codes in the Houston area

We have 5 digits to form

77 is the first two digits ( this is fixed)

For the third digit, we are selecting 1 number out of 0,3,4 or 5

This means 4 C 1

The remaining digits can be any digits

We have 0-9, a total of 10 digits

The first will be 10 C 1 and the second last digit too is 10 C 1

So the number of possible zip codes will be;

4 C 1 * 10 C 1 * 10 C 1

= 4 * 10 * 10 = 400 possible zip codes

6 0

2 years ago