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2 years ago

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If three-fifths of seventh-grade class has missed at least one day of school, what percentage of 7th grade has missed at least a

day of school?

1 answer:

jeyben [28]2 years ago

5 0

Cronchy monchy
bonkier wonkiers 89%

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In general, when a number written in scientific notation includes a positive exponent, the magnitude of the number is __________

Masteriza [31]

Scientific notation is als<span>o </span>called power-of-10 notation, is a method of writing extremely large and small numbers.

when a number written in scientific notation includes a positive exponent, the magnitude of the number is <span>greater than one</span>

8 0

3 years ago

Read 2 more answers

Two containers, one for 3 liters and one for 5 liters, how would you measure 4 liters

Mashutka [201]

Fill the 5-liter container, pour water from that into the 3 liter container until that is full, You will now have 2 liters left in the 5 liter container.

Empty the 3-liter container, and then transfer the 2 liters from the 5-liter container into it.

Now fill the 5-liter container again, then pour water carefully from the 5-liter container into the 3-liter container until it is full - exactly one more liter.

The 5-liter container now has 4 liters in it.

8 0

2 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

3 years ago

What is the product 3x^5(2x^2 +4x+1)

juin [17]

Answer:

6x^7 + 12x^6 + 3x^5

Step-by-step explanation:

Simplify the expression.

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2 years ago

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Helpppppp me plssss<br> uh

Maksim231197 [3]

Answer:

Step-by-step explanation:

-n+1

4 0

2 years ago

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