Which equation is equal to 4a+2b=10

If I add 9 to my number, I get 6. What is my number?

OverLord2011 [107]

negative 3...............................

3 0

3 years ago

What is the exponential form of the logarithmic equation? 3=log0.6 0.216

tatuchka [14]

Answer:

0.216 = 0.6³

Step-by-step explanation:

using the rule of logarithms

• $log_{b}$ x = n ⇔ x = $b^{n}$

$log_{0.6}$ 0.216 = 3 ⇒ 0.216 = $0.6^{3}$

3 0

2 years ago

Read 2 more answers

Anybody know the answer to these question

sashaice [31]

Answer:

No sorry I don't

I did that last year and I don't remember how

Step-by-step explanation:

8 0

2 years ago

For each of the following, determine the interest rate per compounding period. 1) 12% per year, compounded weekly b) 8% per year

Zarrin [17]

Using compound interest, the rates per compounding period are given as follows:

a) 0.1273 = 12.73%.

b) 0.0833 = 8.33%

c) 0.0617 = 6.17%

<h3>What is compound interest?</h3>

The amount of money earned, in compound interest, after t years, is given by:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

In which:

A(t) is the amount of money after t years.

P is the principal(the initial sum of money).

r is the interest rate(as a decimal value).

n is the number of times that interest is compounded per year.

The <u>interest rate per compounding period</u> is given as follows:

$\left(1 + \frac{r}{n}\right)^n - 1$

For item a, the parameters are:

r = 0.12, n = 52.

Hence:

$\left(1 + \frac{0.12}{52}\right)^{52} - 1 = 0.1273$

For item b, the parameters are:

r = 0.08, n = 104.

Hence:

$\left(1 + \frac{0.08}{104}\right)^{104} - 1 = 0.0833$

For item c, the parameters are:

r = 0.06, n = 12.

Hence:

$\left(1 + \frac{0.06}{12}\right)^{12} - 1 = 0.0617$

More can be learned about compound interest at brainly.com/question/25781328

#SPJ1

8 0

1 year ago

Determine whether the integral is convergent or divergent. [infinity] 7 e−1/x x2 dx convergent divergent Changed: Your submitted

inessss [21]

I suppose the integral could be

$\displaystyle\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx$

In that case, since $-\frac1x\to0$ as $x\to\infty$ , we know $e^{-1/x}\to1$ . We also have $\left(e^{-1/x}\right)'=\frac{e^{-1/x}}{x^2}>0$ , so the integral is approach +1 from below. This tells us that, by comparison,

$\displaystyle\frac{e^{-1/x}}{x^2}\le\frac1{x^2}\implies\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx\le\int_7^\infty\frac{\mathrm dx}{x^2}$

and the latter integral is convergent, so this integral must converge.

To find its value, let $u=-\frac1x$ , so that $\mathrm du=\frac{\mathrm dx}{x^2}$ . Then the integral is equal to

$\displaystyle\int_{-1/7}^0e^u\,\mathrm du=e^0-e^{-1/7}=1-\frac1{\sqrt[7]{e}}$

4 0

3 years ago