Answer:
Sheridan's Work is correct
Step-by-step explanation:
we know that
The lengths side of a right triangle must satisfy the Pythagoras Theorem

where
a and b are the legs
c is the hypotenuse (the greater side)
In this problem
Let

substitute

Solve for b





we have that
<em>Jayden's Work</em>


substitute and solve for c





Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle
<em>Sheridan's Work</em>


substitute

Solve for b





therefore
Sheridan's Work is correct
Answer:
z=40 degrees
Step-by-step explanation:
a triangle is 180 degrees
<ACB=50 (180-130=50)
50+90=140
180-140=40
Answer:
B
Step-by-step explanation:
The answer is B because if a line goes through (0,0) then it is automatically proportional.
Divide 296 by 16 you get 18.5 then add the 5.75 to get 24.25
So the answer is 24.25
Answer:

Step-by-step explanation:
A polynomial is given to us and we need to make the polynomial by its degree . Here the given polynomial is 3x² + 5x -3 .

So here we can see that the highest power of the variable in the given polynomial is 2 . So the polynomial is a quadratic polynomial .
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When we equate this quadratic polynomial with 0 , it becomes a quadratic equation.
That is ,
This is done in order to find the zeroes of the polynomial. The zeroes of the polynomial are the values for which the polynomial becomes 0 .