Answer:
Step-by-step explanation:
Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.
Problem 1:
(x²)/4 +y²= 1
y= x+1
*substitute for y*
Now we have a one-variable equation we can solve-
x²/4 + (x+1)² = 1
x²/4 + (x+1)(x+1)= 1
x²/4 + x²+2x+1= 1
*subtract 1 from both sides to set equal to 0*
x²/4 +x^2+2x=0
x²/4 can also be 1/4 * x²
1/4 * x² +1*x² +2x = 0
*combine like terms*
5/4 * x^2+2x+ 0 =0
now, you can use the quadratic equation to solve for x
a= 5/4
b= 2
c=0
the syntax on this will be rough, but I'll do my best...
x= (-b ± √(b²-4ac))/(2a)
x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))
x= (-2 ±√(4-0))/(2.5)
x= (-2±2)/2.5
x will have 2 answers because of ±
x= 0 or x= 1.6
now plug that back into one of the equations and solve.
y= 0+1 = 1
y= 1.6+1= 2.6
Hopefully this explanation was enough to help you solve problem 2.
Problem 2:
x² + y² -16y +39= 0
y²- x² -9= 0
Answer:
Part 1) 
Part 2) 
Part 3) 
Step-by-step explanation:
we know that
The old account balance plus the transaction amount is equal to the new account balance
The transaction amount can be a positive number (example a deposit) or a negative number (example a withdrawal)
Part 1) Find the value of A

solve for A
subtract 432 both sides

Part 2) Find the value of B

solve for B

Rewrite

Part 3) Find the value of C

solve for C
subtract 52 both sides


Answer: 4.25 pound of meat
Step-by-step explanation:
This is a addition questions, all you have to do is to add the numbers together. But you have to convert all the fractions and decimal to the same thing first. In this case, I'll convert all the fraction to decimal.
2.42 + 0.75 + 0.36 ✕ 3 = 4.25 pound
Answer:
20.48
The parentheses just means multiplication.
Answer:
≈50.6
Step-by-step explanation:
Not sure what precision level this problem is looking for, but for right-skewed distributions, we know that the mean is going to be pulled right and therefore the mean should be larger than the median. To a high confidence level, the mean should fall between 50 and 59, or in the third column.
If a single estimation is wanted, assume the values inside each column are uniformly distributed:
