Ask question

Ask question

All categories

3 years ago

10

What are the solutions of the equation x^4-5x^2-36=0? Use factoring to solve

1 answer:

Murljashka [212]3 years ago

3 0

X = -3, 3, 2i, -2i
Hope this helps! :)

You might be interested in

Find the indicated angle . (Use either the Law of Sines or the Law of Cosines, as appropriate. Assume a = 16, c = 12, and ∠C = 4

mote1985 [20]

Use the law of sines
11/sin50°=13/sinθ
11/0.766=13/sinθ
11sinθ=13(0.766)
11sinθ=9.958
sinθ=9.958/11
sinθ=0.9052727
use arcsin on your calculator
θ=64.86°

7 0

2 years ago

Please help me with 8 im so very stuck i know how to do those but nothing will factor into eachother

Kisachek [45]

Answer:

lol number 8 is x= i, -i , $\frac{3}{4}$ , -4 so u are correct.

and number 7 is x= $-\frac{1}{2}$ or $\sqrt[5-i]\sqr\frac{23}{8}$

Step-by-step explanation:

7 0

3 years ago

One spring day, Landon noted the time of day and the temperature, in degrees Fahrenheit. His findings are as follows: At 6 a.m.,

Verizon [17]

Answer:

Graph these points listed below

Step-by-step explanation:

Graph the following points:

(6 am, 50)

(9 am, 59)

(1 pm, 63)

(6 pm, 63)

(8 pm, 59)

(12 am midnight, 58)

6 0

3 years ago

4 sailboats to 12 motorboats

lukranit [14]

Answer:

4:12

Step-by-step explanation:

I think you asked for a ratio?

8 0

3 years ago

Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$

Simplify the radical with the cube:

$\displaystyle x^3=\frac{3y+16}{2y+9}$

Multiply by 2y+9

$\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)$

Simplify:

$\displaystyle x^3(2y+9)=3y+16$

Operate the parentheses:

$\displaystyle x^3(2y)+x^3(9)=3y+16$

$\displaystyle 2x^3y+9x^3=3y+16$

Subtract 3y and $9x^3$ :

$\displaystyle 2x^3y - 3y=16-9x^3$

Factor y out of the left side:

$\displaystyle y(2x^3 - 3)=16-9x^3$

Divide by $2x^3 - 3$ :

$\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}$

ii) To find y when x=2, substitute:

$\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}$

$\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}$

$\displaystyle y=\frac{16-72}{16- 3}$

$\displaystyle y=\frac{-56}{13}$

$\mathbf{\displaystyle y=-\frac{56}{13}}$

8 0

3 years ago