Cannot guarantee. But I believe the answers are:
1) (3,2)
2) (3,-2)
3) (-3,2)
4) (-3,-2)
Edit: These are correct. When you substitute these into each equation,you get 72.
The negatives work because (-x)^2 = positive
Using online calculators is great
Short Answer 5/216
Comment
The question really is, how many different types of throws with 4 dice will give 21? You can count them
Pattern 1
6663
Pattern 2
5556
Pattern 3 [ The hard one]
6654
Patterns 1 and 2 give 4 each.
6 6 6 3
6 6 3 6
6 3 6 6
3 6 6 6
5553 will do the same thing
Both can be found by (4/1) as a combination.
Now for 6654 That gives 12
6 6 5 4
6 6 4 5
6 5 4 6
6 5 6 4
6 4 5 6
6 4 6 5
5 6 4 6
5 6 6 4
5 4 6 6
4 5 6 6
4 6 5 6
4 6 6 5
That should total 12
The total number of ways of getting 21 with this pattern is 12
Total successes
12 + 4 + 4 = 30
What is the total number of ways you can throw 4 dice?
Total = 6 * 6 * 6 * 6 = 1296
What is the probability of success?
30 / 1296 = 5 / 216
Answer:
v = 7
is the value for which
x = (-21 - √301)/10
is a solution to the quadratic equation
5x² + 21x + v = 0
Step-by-step explanation:
Given that
x = (-21 - √301)/10 .....................(1)
is a root of the quadratic equation
5x² + 21x + v = 0 ........................(2)
We want to find the value of v foe which the equation is true.
Consider the quadratic formula
x = [-b ± √(b² - 4av)]/2a ..................(3)
Comparing (3) with (2), notice that
b = 21
2a = 10
=> a = 10/2 = 5
and
b² - 4av = 301
=> 21² - 4(5)v = 301
-20v = 301 - 441
-20v = -140
v = -140/(-20)
v = 7
That is a = 5, b = 21, and v = 7
The equation is then
5x² + 21x + 7 = 0