What's .77 divided by 236.9 ?

Determine whether quantities vary directly or inversely and find the constant of variation. A teacher grades 25 students essays

Lesechka [4]

IDK the constant and quantities...
First, find how many hours it takes for him to grade 1 essay, 0.16. Then multiply that by 35 to get 5.6.
I hope this is right -_-'

8 0

3 years ago

Expand and simplify 3(2a+5)+5(a-2)

Wittaler [7]

Step 1 (3* (2a+5)) + 5 *(a-2)
Step 2 <span>3 • (2a + 5) + 5 • (a - 2)
</span>Step 3 final result and your anwser
11a+5
Hope it helped if it did it that tanks button for me tanks

3 0

3 years ago

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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

Jesus bought 5/6 pounds of peanuts. He ate 3/4 pound of the peanuts with his friends. how much does Jesus have left?

Bad White [126]

Answer:

Step-by-step explanation:

Total peanuts = 5/6pounds

They ate = 3/4 of the peanut

= 3/4 × 5/6 = 5/8pounds

They eat 5/8 pounds

Amount left = 5/6 - 5/8

= (20 - 15)/24

= 5/24pounds

Jesus had 5/24 pounds of peanuts left

5 0

3 years ago

Ik it’s 7 in the morning but can somebody help a gal out

faust18 [17]

Answer:

10

Step-by-step explanation:

The vertical scale gives us the frequency or the number of pets under each category . From the diagram;

The number of Dogs is 11 while there is only 1 horse

The difference between this numbers will be the solution to the question posed;

11 - 1 = 10

Therefore, 10 more dogs are pets than horses

4 0

4 years ago

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