Answer:
Using a Graph of a Line Identify the x-axis. A coordinate graph has a y-axis and an x-axis. The x …
Using the Equation of the Line Determine that the equation of the line is in standard form. The …
Using the Quadratic Formula Determine that the equation of the line is a quadratic equation. A
Step-by-step explanation:
What you have to do is add them all!
20 bar
11.4 2 11.4 weights
11.4
4.5 2 4.5 weights
4.5
450.0 4 450 weights
450.0
450.0
+450.0
---------
1.8
10
200
+1,600
--------------
1,811.8 is the answer
Answer:
no solutions
Step-by-step explanation:
Hi there!
We're given this system of equations:
x+y=3
4y=-4x-4
and we need to solve it (find the point where the lines intersect, as these are linear equations)
let's solve this system by substitution, where we will set one variable equal to an expression containing the other variable, substitute that expression to solve for the variable the expression contains, and then use the value of the solved variable to find the value of the first variable
we'll use the second equation (4y=-4x-4), as there is already only one variable on one side of the equation. Every number is multiplied by 4, so we'll divide both sides by 4
y=-x-1
now we have y set as an expression containing x
substitute -x-1 as y in x+y=3 to solve for x
x+-x-1=3
combine like terms
-1=3
This statement is untrue, meaning that the lines x+y=3 and 4y=-4x-4 won't intersect.
Therefore the answer is no solutions
Hope this helps! :)
The graph below shows the two equations graphed; they are parallel, which means they will never intersect. If they don't intersect, there's no common solution
Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

We have
and
, because the dice are fair.
Now we use the assumption of independence to claim that

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:
- 2 in a unique way (1+1)
- 3 in two possible ways (1+2, 2+1)
- 4 in three possible ways
- 5 in three possible ways
- 6 in three possible ways
- 7 in two possible ways
- 8 in a unique way
This implies that the probabilities of the outcomes of
are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5