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Oksana_A [137]
2 years ago
11

Please tell me the answer and explain how you got it

Mathematics
1 answer:
Charra [1.4K]2 years ago
8 0
Which answer :) xx just for help
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Slope of -1,0 and 10,18
steposvetlana [31]

You should find the slope using (y2 - y1) / (x2 - x1) = m

m = (18 - 0) / (10 - (-1))

m= 18 / 11

Therefore the slope is 18/11

6 0
3 years ago
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What is the approximate volume of the cone? use 3.14 for
Natali5045456 [20]

Answer:

157 cm²

Step-by-step explanation:

volume = (1/3) * π * r² * h

volume = (1/3) * π * 5² * 6

volume = (1/3) * π * 25 * 6

volume = (1/3) * π * 150

volume = (1/3) * 471

volume = 157 cm²

8 0
3 years ago
It cost Faith $9.10 to send 91 text messages. How many text messages did she send if she spent $17.60?
Komok [63]

Answer:

um

Step-by-step explanation:

176 10cent per 1 message

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2 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
Which value for y makes the equation y - 2.5 = 1.95 true? A. y = 0.78 B. y = 4.875 c. y = 48.75 D. y = 4,875​
irina1246 [14]

Answer: B is the answer

Step-by-step explanation:

7 0
2 years ago
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