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2 years ago

7

How to differentiate a function?

1 answer:

kiruha [24]2 years ago

5 0

Answer:

I have a formula for differentiating a questions on differentiation

Step-by-step explanation:

I have not published it yet but can give a brief summary of it will later use it to solve questions here after publishing it

any time you a given a function or a question to differentiate always write it in this form y=am...will explain the rest later after publishing it. it works for all kind of differentiation questions like product or quotient type of questions

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Pls answer and show WORK pls do it right

adelina 88 [10]

Answer:

4/6

Step-by-step explanation:

1/2 needs to be converted to sixths, so we would multiple the numerator and the denominator by 3, because 2 goes into 6, 3 times, then we would get 3/6 + 1/6, then just add, 1 + 3 = 4, and then you end up with 4/6. Have A Great Day!

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3 years ago

63/40 divided by 70/44

const2013 [10]

Answer:

0.90

Step-by-step explanation:

I entered it onto my calculator, but if you'd like an explanation as to how just ask.

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3 years ago

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In a function of the form y= ax^2 +bx + c which of the following is always true? A. If a is positive, the parabola will have a m

GuDViN [60]

A is always true

If a is negative the parabola wil have a minimum value

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3 years ago

PLZ HELP ASAP!!!!<br> MARK BRAINLIEST!!!

lesantik [10]

Answer:

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Step-by-step explanation:

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2 years ago

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Use the compound interest formulas A= P (1 + r/n) ^nt and A= Pe ^ rt to solve the problem given. Round to the nearest cent. Find

Dennis_Churaev [7]

Answer:

$\$30,460$
$\$30,497$
$\$30,522$
$\$30,535$

Step-by-step explanation:

We know that,

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

where,

A = Amount after time t,

P = Principle amount,

r = Rate of interest,

n = Number of times interest is compounded per year,

t = time period in year.

Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded semiannually

Here,

P = $25,000

r = 5% = 0.05

n = 2 (as compounded semiannually)

t = 4 years

Putting the values,

$A=25000\left (1+\dfrac{0.05}{2}\right )^{2\times 4}$

$=25000\left (1+0.025\right )^{8}$

$=25000\left (1.025\right )^{8}$

$=\$30,460$

Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded quarterly.

Here,

P = $25,000

r = 5% = 0.05

n = 4 (as compounded quarterly)

t = 4 years

Putting the values,

$A=25000\left (1+\dfrac{0.05}{4}\right )^{4\times 4}$

$=25000\left (1+0.0125\right )^{16}$

$=25000\left (1.0125\right )^{8}$

$=\$30,497$

Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded monthly.

Here,

P = $25,000

r = 5% = 0.05

n = 12 (as compounded monthly)

t = 4 years

Putting the values,

$A=25000\left (1+\dfrac{0.05}{12}\right )^{12\times 4}$

$A=25000\left (1+\dfrac{0.05}{12}\right )^{48}$

$=\$30,522$

Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded continuously.

$A= Pe^{rt}$

where,

A = Amount after time t,

P = Principle amount,

r = Rate of interest,

t = time period in year.

Putting all the values,

$A= 25000e^{0.05\times 4}=\$30,535$

It can be observed that, the frequent we compound the amount, the more we get.

7 0

2 years ago