Let the no be X and Y

acc to ques....

x-y=9 .........1

xy=162 ..........2

substituting value from 1 in 2 we get;

x=9+y

[9+y][y] = 162

y^2+9y = 162

y^2 + 9y - 162=0

y^2 + 18y - 9y - 162=0

y[y+18] + 9[y+18]=0

[y+9][y+18}

y= -9.................................3

y= -18......................................4

case 1 :

y= -9

x = 9-9=0

case 2:

y= -18

x= 9-18 = -9

7 0

2 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

5 months ago

In the function y=4x^2-5, what effect does the number -5 have on the graph, as compared to the graph of y=x^2

kati45 [8]

It moves the graph 5 steps downwards on the y axis.

3 0

3 years ago

Solve the equation. -(7-4x)=9

Ket [755]

Answer:

x=4

Step-by-step explanation:

-(7-4x)=9

-7+4x=9

4x=16

x=4

7 0

2 years ago

Read 2 more answers