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3 years ago

WILL MAKE BRAINLIEST!!

Mathematics

2 answers:

qwelly [4]3 years ago

7 0

Answer:

(a) Jun's work was correct as far as it went, but she didn't go far enough.

(b) (1, 5) and (-4, 30)

Step-by-step explanation:

y = x² + 10x – 6

y = 7x – 2

(a) Jun’s solution:

y = x² + 10x – 6

y = 7x – 2

x² + 10x – 6 = 7x – 2

x² + 3x – 4 = 0

(x – 1)(x + 4) = 0

x = 1 or x = -4

Jun's work was correct <em>to this point</em>, but <em>she didn't go far enough</em>. She calculated the x-intercepts of the combined equation, not the values of x and y at which the two equations intersect.

Jun must calculate the value of y that corresponds to each value of x.

(b) The correct solution

Continue Jun's calculations

y = 7x -2 y = 7x - 2

Set x = 1 Set x = -4

y = 7 - 2 y = -28 -2

y = 5 y = -30

The correct solutions are (1, 5) and (-4, 30).

The Figure below shows the graphs of each function with the intersection points at (1, 5) and (-4, 30).

Naddika [18.5K]3 years ago

5 0

Answer:

B. juans solution is incorrect

Step-by-step explanation:

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Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2

Basile [38]

The Laplace transform of the given initial-value problem

$y' 5y = e^{4t}, y(0) = 2$ is mathematically given as

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is mathematically given as

$&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}$

$\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}$

$put $s=-s \Rightarrow a_{1}=\frac{17}{9}$$

$\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}$

In conclusion, Taking inverse Laplace tranoform

$L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\$

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$