You have to write two equation:
'substract 22 from a number'

and:
'the result is doubled, the answer is 6 more than the original number'

then you change 'Y' in 2nd equation for left side of 1st equation:

count it and you get:

This means that X=50
The original number is 50.
The solution for x in the equation 3ax - n/5 = -4 is x = -4/3a + n/15a.
<h3>What is the solution for x in the given equation?</h3>
3ax - n/5 = -4
To solve for x, isolate the term that contains x.
3ax - n/5 = -4
3ax = -4 + n/5
Divide each term by the coefficient of x and simplify
3ax/3a = -4/3a + (n/5)/3a
x = -4/3a + (n/5)/3a
multiply (n/5) by the reciprocal of 3a
x = -4/3a + (n/5) × 1/3a
x = -4/3a + n/15a
Therefore, the solution for x in the equation 3ax - n/5 = -4 is x = -4/3a + n/15a.
Learn more about equations here: brainly.com/question/14686792
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Answer:
The perimeter of the polygon is 
Step-by-step explanation:
we know that
The perimeter is the sum of the length sides of the polygon
In this problem
P=XY+YZ+XZ
the formula to calculate the distance between two points is equal to
step 1
Find the distance XY
we have
X(−1, 3), Y(3, 0)
substitute
step 2
Find the distance YZ
we have
Y(3, 0), Z(−1,−2)
substitute
step 3
Find the distance XZ
we have
X(−1, 3), Z(−1,−2)
substitute
step 4
Find the perimeter
P=XY+YZ+XZ

So, to evaluate a combination, there's a formula we use.
I don't remember the formula from the top of my head, lol, but this is how you solve them.
7 c 2
When doing combinations and permutations each number is always in a factorial. We always start with the number on the left.
7! That's the total amount. The number on the left divides into that.
7! / 2!
We're not done yet. Here's the tricky part. We also always divide the number on the left, in this case 7!, with the positive difference of both numbers given to us.
7 - 2 = 5
So, we have 7! / 5! / 2! = 21.
Hope that helped!
Let's work another one.
5 c 3
We have 5! / 3! ,but we need to also divide 5! by the positive difference of 5 and 3. We get 2.
So, 5! / 3! / 2! = 10.
If you have any questions then leave a comment. Good luck!
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