Solve the equation `x^{4}+13x^{2}+36=0` for all values of `x`.

Mathematics

1 answer:

Zolol [24]3 years ago

7 0

Answer: x4+13x2+36=0

Step-by-step explanation:

Let's solve your equation step-by-step.

x4+13x2+36=0

Step 1: Factor left side of equation.

(x2+4)(x2+9)=0

Step 2: Set factors equal to 0.

x2+4=0 or x2+9=0

You might be interested in

The tickets in a box are numbered from 1-20 inclusive. If a ticket is chosen at random and replaced, and then a second ticket is

ozzi

7 + 5 = 12 so is an even number that the ticket the choose

3 0

3 years ago

The mass of Hinto's math is 4,367 grams. What is the mass of 5 math books in kilograms?

suter [353]

Answer:

21.835 kilograms

Step-by-step explanation:

(4,367x5)/1000

7 0

3 years ago

AB is perpendicular to CD. How many 90 degree angles are formed by the intersection

Alex73 [517]

It should be 4 90°angles

8 0

3 years ago

Read 2 more answers

Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}$

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0

3 years ago

Can someone answer this for me? I don’t understand how to do i.

LiRa [457]

These are the steps good luck

7 0

3 years ago