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kherson [118]
3 years ago
11

Which is the solution to the system of equations?

Mathematics
1 answer:
Ivenika [448]3 years ago
4 0

Answer:

D. (2, -3/4)

Step-by-step explanation:

y=1/8x-1

-5x+4y=-13

5x+4(1/8x-1)=-13

x=2

y=1/8x2-1

y=-3/4

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Find an equation that models the path of a satellite if its path is a hyperbola, a = 55,000 km, and c= 81,000 km. Assume the cen
elena-14-01-66 [18.8K]

Answer:

\frac{x^2}{55000^2} - \frac{y^2}{59464^2} =1

Step-by-step explanation:

the transverse axis is horizontal.

so its a horizontal hyperbola

Center is the origin so center is (0,0)

Equation of horizontal hyperbola is

\frac{x^2}{a^2} - \frac{y^2}{b^2} =1

Given a= 55000 and c= 81000

c^2 = a^2 + b^2

81000^2 = 55000^2 + b^2

subtract 55000^2 on both sides

b  = sqrt(81000^2 - 55000^2)= 59464.27

now plug in the values

\frac{x^2}{55000^2} - \frac{y^2}{59464^2} =1

7 0
2 years ago
Blake is driving home from the water park. He drives for 60 miles and stops to rest before driving 20 more miles. Unfortunately,
7nadin3 [17]
Mathematically its 60+20-36+70
7 0
3 years ago
Read 2 more answers
This is due tm please help​
Mumz [18]

Answer:

25. No slope

Step-by-step explanation:

25. Y2 - Y1 / X2 - X1

= -8 - (-3) / 9 - 9

= -5 / 0

= no slope; straight line

Use this formula (Y2 - Y1 / X2 - X1) for each ques.

8 0
3 years ago
Whats the lowest common multiple of 120 and 19600​
Levart [38]

Answer:

Multiples of 120 are 120, 240, 360, 480, 600, 720, 840 etc; Multiples of 150 are 150, 300, 450, 600, 750, 900 etc; Therefore, the least common multiple of 120 and 150 is 600.

Least common multiple (LCM) of 19600 and 19619 is 384532400.

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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
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