I need to know the equation for this table!

Mathematics

1 answer:

Kobotan [32]3 years ago

7 0

The equation for the table is y=2x+3.

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What are the solutions of the equation x^4 + 6x^2 + 5 = 0? Use u substitution to solve.

Radda [10]

Answer:

second option

Step-by-step explanation:

Given

$x^{4}$ + 6x² + 5 = 0

let u = x², then

u² + 6u + 5 = 0 ← in standard form

(u + 1)(u + 5) = 0 ← in factored form

Equate each factor to zero and solve for u

u + 1 = 0 ⇒ u = - 1

u + 5 = 0 ⇒ u = - 5

Change u back into terms of x, that is

x² = - 1 ( take the square root of both sides )

x = ± $\sqrt{-1}$ = ± i ( noting that $\sqrt{-1}$ = i ), and

x² = - 5 ( take the square root of both sides )

x = ± $\sqrt{-5}$ = ± $\sqrt{5(-1)}$ = ± $\sqrt{5}$ × $\sqrt{-1}$ = ± i $\sqrt{5}$

Solutions are x = ± i and x = ± i $\sqrt{5}$

3 0

2 years ago

20 POINTS!!!

notka56 [123]

$\dfrac{2}{\sqrt3\cos x+\sin x}=\sec\left(\dfrac{\pi}{6}-x\right)\\\\\dfrac{2}{\sqrt3\cos x+\sin x}=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}\ \ \ \ \ (*)\\----------------\\\\\cos\left(\dfrac{\pi}{6}-x\right)\ \ \ \ |\text{use}\ \cos(x-y)=\cos x\cos y+\sin x\sin y\\\\=\cos\dfrac{\pi}{6}\cos x+\sin\dfrac{\pi}{6}\sin x=\dfrac{\sqrt3}{2}\cos x+\dfrac{1}{2}\sin x=\dfrac{\sqrt3\cos x+\sin x}{2}\\------------------------------$

$(*)\\R_s=\sec\left(\dfrac{\pi}{6}-x\right)=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}=\dfrac{1}{\dfrac{\sqrt3\cos x+\sin x}{2}}\\\\=\dfrac{2}{\sqrt3\cos x+\sin x}=L_s$