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3 years ago

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.

4 0

2 years ago

Read 2 more answers

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