Consider the equation
.
First, you can use the substitution
, then
and equation becomes
. This equation is quadratic, so
.
Then you can factor this equation:
.
Use the made substitution again:
.
You have in each brackets the expression like
that is equal to
. Thus,
![x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)](https://tex.z-dn.net/?f=%20x%5E3%2B5%3D%28x%2B%5Csqrt%5B3%5D%7B5%7D%29%28x%5E2-%5Csqrt%5B3%5D%7B5%7Dx%2B%5Csqrt%5B3%5D%7B25%7D%29%20%2C%5C%5Cx%5E3%2B1%3D%28x%2B1%29%28x%5E2-x%2B1%29%20%20%20)
and the equation is
.
Here
and you can sheck whether quadratic trinomials have real roots:
1.
.
2.
.
This means that quadratic trinomials don't have real roots.
Answer:
If you need complex roots, then
.
When two balanced dice are rolled, the sum of the dice can be 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12, giving 11 possibilities. th
RUDIKE [14]
False.
There are 36 possible outcomes, corresponding to numbers 1-6 independently appearing on each of the dice. Only one of those outcomes is double-sixes, resulting in a sum of 12. The probability that the sum is 12 is 1/36.
In short, the 11 outcomes listed in your problem statement are not equally-likely.
Answer:325,000
Step-by-step explanation:
4 gets rounded up to 5 because of the 6 next to it.
Answer:
the answer is b i click it and it was right
y=2(3)^x
Step-by-step explanation: