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MariettaO [177]

3 years ago

9

using compound interest, how long does it take to double a 1000 dollar investment that pays 6.5% annual interest, compounded mon

thly?

1 answer:

KonstantinChe [14]3 years ago

3 0

Answer:

11 years approx

Step-by-step explanation:

Given data

P=$1000

A=2000

R=6.5%

T= ?

Calculate time, solve for t

t = ln(A/P) / r

substitute

t=ln(2000/1000)/0.065

t=ln(2)/0.065

t=0.693/0.065

t=10.66

Hence the time is 11 years approx

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The table shows a linear function.

sergejj [24]

Answer:

$\large\boxed{y=\dfrac{5}{3}x+9}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

<em>m</em><em> - slope</em>

<em>b</em><em> - y-intercept → (0, b)</em>

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

From the table we have an y-intercept (0, 9) → b = 9.

Take other point from the table (-6, -1).

Calculate the slope:

$m=\dfrac{-1-9}{-6-0}=\dfrac{-10}{-6}=\dfrac{5}{3}$

Finally:

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7 0

3 years ago

Please help! Thanks in advance.

alukav5142 [94]

In this problem, we need to plug in the given x values for $f(x)=0$ and find a and b.

When we plug in 1, we get: $2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0$

Simplify:
$2 - 5 - 14 + a + b = 0$
$- 17 + a + b = 0$
$a + b = 17$

We got our first statement about the values of the variables. If we find one more we can find those 2 variables.

We have another given root: 4.

Plug it in:
$2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0$
$512 - 320 - 224 + 4a + b = 0$
$- 32 + 4a + b = 0$
$4a + b = 32$

Now we have our second one. We can combine them:
$a + b = 17 \\ 4a + b = 32$

I use elimination method which is easier here.

Multiply the top equation by -1: $- a - b = - 17 \\ 4a + b = 32$

Add them up:
$3a = 15$

Simplify:
$a = 5$

Now we have a, we can plug in one of those equations to find b:
$5 + b = 17$
$b = 17 - 5$
$b = 12$

So, the answers are $a=5$ and $b=12$ .

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Pendejo Que putas es eso

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Answer:

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Step-by-step explanation:

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