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2 years ago

In this arithmetic sequence, what is the common difference?

Mathematics

2 answers:

sergiy2304 [10]2 years ago

8 0

They are subtracting 3 from each previous term, so the answer is B: -3. You're welcome. ;)

-Dominant- [34]2 years ago

5 0

The answer is -3 because you are subtracting 3 every time.

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A circle is inscribed in a square, as shown in the figure below. The length of

mafiozo [28]

Answer:

78.5 cm^2

Step-by-step explanation:

the circle's diameter is 10cm --> its radius is 5cm

the area of the circle is 5^2 x 3.14 = 78.5cm^2

4 0

3 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

2 years ago

Carlo buys $14.40 worth of grapefruit. Each grapefruit costs $0.80.<br> how many does he have

gtnhenbr [62]

Answer:

Mate, just divide...its that simple..

$14.40 divided by 0.80= 18

ANSWER: 18

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Mary states, If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. Decide if her statement i

Mila [183]

The answer for apex would be true.

8 0

3 years ago

Read 2 more answers

Help help please help help

yuradex [85]

Answer:

The range is 5.

Step-by-step explanation:

You can deduce this by concluding all other statements are true, or by calculating the range. 10 - 1 = 9

3 0

3 years ago