Sum of 2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=

for x²-4x+16=0
x=

x=

x=

x=

x=

x=

the roots are
x=-4 and 2+2i√3 and 2-2i√3
Answer:
<u>7</u><u>.</u><u>7</u><u> </u><u> </u>= <u>2</u><u>.</u><u>3</u>
3.6 b
<u>7</u><u>.</u><u>7b</u>= <u>8</u><u>.</u><u>28</u>
7.7 7.7
b = 1.075324675
b = 1.08
A) If the last digit in the fractional part of 1.080 is less than 5, then simply remove the last the digit of the fractional part. With 1.080, rule A applies and 1.08 rounded to the nearest hundredth is:
No, Lonnie is not correct because the correct answer is 5/6. If you do 35/42 divide both by 7, it'll be 5/6.
Step-by-step explanation:
Use the first fundamental theorem of calculus.
∫₆¹⁰ f'(x) dx = f(10) − f(6) = 8 − 8 = 0