Suppose p varies directly as the cube root of q. If p=-7 when q=11, what is P if q is 12? Help!!!?

p varies directly as the cube root of q

p / ³√q is a constant.

p = -7 when q = 11. What is p when q is 12?

-7 / ³√11 = p / ³√12

p = -7 ³√(12/11)

p = k(³√q) → ³√q = p/k → q = p³/k³ → 11k³ = -343 and therefore k³ = -343/11

hence q = 12 = 11p³/-343 giving p³ = -12 x 343/11, so p = -7.206........

p = k * q^(1/3)

If p=-7 when q=11, then k = -7/11^(1/3)

When q = 12, p = -7/11^(1/3) * 12^(1/3)

So you have ~ q^(1/3) if you intorduce a constant of proportionality, k, then you can write

p = k q^(-1/3)

Now p = -7 when q = 11 so --> k = p/q^(1/3) = -7/(11)^(1/3) = -3.1475

Nlw set q = 12 ---> p = -3.1475*(12)^(1/3) = -7.206