It depends how much she begins with but if she doesn’t begin with any then it’d be 2.5
Answer:
4(k - 3)(3k + 5)
Step-by-step explanation:
Given
12k² - 16k - 60 ← factor out 4 from each term
= 4(3k² - 4k - 15) ← factor the quadratic
Consider the factors of the product of the coefficient of the k² term and the constant term which sum to give the coefficient of the k term
product = 3 × - 15 = - 45 , sum = - 4
Factors are - 9 and + 5
Use these factors to split the middle term
3k² - 9k + 5k - 15 → ( factor the first/second and third/fourth terms
= 3k(k - 3) + 5(k - 3) ← factor out (k - 3)
= (k - 3)(3k + 5)
Hence
12k² - 16k - 60 = 4(k - 3)(3k + 5) ← in factored form
Answer:
y=mx+b :)
Step-by-step explanation:
Answer:
<h2>3(cos 336 + i sin 336)</h2>
Step-by-step explanation:
Fifth root of 243 = 3,
Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),
then r^5( cos Ф + i sin Ф )^5 = 243(cos 240 + i sin 240).
Equating equal parts and using de Moivre's theorem:
r^5 =243 and cos 5Ф + i sin 5Ф = cos 240 + i sin 240
r = 3 and 5Ф = 240 +360p so Ф = 48 + 72p
So Ф = 48, 120, 192, 264, 336 for 48 ≤ Ф < 360
So there are 5 distinct solutions given by:
3(cos 48 + i sin 48),
3(cos 120 + i sin 120),
3(cos 192 + i sin 192),
3(cos 264 + i sin 264),
3(cos 336 + i sin 336)
