All categories

4 years ago

Find the direct relationship between y and x . x=e^2t y=e^t

Mathematics

2 answers:

nekit [7.7K]4 years ago

8 0

Answer:

Step-by-step explanation:

x=e^2t

y=e^t

you. have : x=e^2t =(e^t)² = y²

the direct relationship between y and x is : x = y²

olya-2409 [2.1K]4 years ago

5 0

Answer:

x = y²

Step-by-step explanation:

Given

x = $e^{2t}$ and y = $e^{t}$

Note that

x = $e^{2t}$ = $e^{t}$ × $e^{t}$ = y²

Thus x = y²

You might be interested in

Write the slope-intercept form of the equation of each line. 9x-7y=-7

makkiz [27]

Slope intercept form is y = mx + b.....so lets solve for y

9x - 7y = -7 .....subtract 9x from both sides
-7y = -9x - 7...now we divide both sides by -7
y = 9/7x + 1 <==

6 0

3 years ago

Which of the following are solutions to this equation 3n+4=24-n a) n=1 b) n=3 c) n=5 d) n=7

g100num [7]

Answer:

The correct option is c) n=5.

Therefore the solution to

$3n+4=24-n$

$n=5$

Step-by-step explanation:

Given:

$3n+4=24-n$

To Find;

n = ?

Solution:

$3n+4=24-n$ ........Given

Add + n to both the side

$3n+4+ n=24-n+n\\4n+4=24$

Now Subtract 4 on both the side

$4n+4-4=24-4\\4n=20$

Dividing by 4 on both side

$\dfrac{4n}{4}=\dfrac{20}{4}\\\\n=5$

Therefore the solution to

$3n+4=24-n$

$n=5$

4 0

3 years ago

Help plsss

LenKa [72]

<h3></h3><h2>⭐ Development:</h2><h3></h3>

Having the expression, let's modify it so it becomes a 2nd degree equation:

$\large {\text {$ \sf \cfrac{1}{2x^2} -2 = 0 $}}$

Now, we will multiply per 2x² both sides of equation...

$\large {\text {$ \sf \cfrac{1}{2x^2}\cdot \:2x^2-2\cdot \:2x^2=0\cdot \:2x^2 $}$

$\searrow$

$\large {\text {$ \sf 1-4x^2=0$}}$

We have to write in standard form...

$\large {\text {$ \sf -4x^2+1 = 0 $}}$

<u />

$\large {\text{$\sf x=\cfrac{-b\pm\sqrt{b^2-4ac}}{2a} \quad\rightarrow\quad x=\cfrac{-0\pm\sqrt{0^2-4\cdot (-4) \cdot1} }{2\cdot (-4) } \:\rightarrow\:\: x=\cfrac{-0^2 \pm4}{2 \cdot(-4)} $}}$

$\huge {\text {$ \sf \downarrow$}}$

$\large {\text {$\sf {\bf x_1} = \cfrac{-0+4}{2\left(-4\right)}= \cfrac{-1}{2} $}}$ $\large {\text {$\sf {\bf x_2 }=\cfrac{-0-4}{2\left(-4\right)} = \cfrac{1}{2} $}}$