4. In the exit slip, Brady got everything wrong. He needed to factor

x2 - 10x + 25 and 16x2 -9. This is his work.

x2 - 10x + 25 = x2 - 10x + 52

= x2 - 2(x)(5) + 52

= (x + 5)(x - 5)

16c2 – 9 = (40)2 – (3)2

= (40 - 3)2

Identify the error(s) in planning the solution or solving the problem. Write

the correct solution.

vo

r

of choice to

S

(1)The error is in third step.

The correct solution is (x-5)²

(2)The error is in third step.

The correct solution is (4c+3)(4c-c)

Step-by-step explanation:

(a-b) whole square is the sum square of a and b and subtraction of product of 2,a and b.

(a-b) = a²+b²-2ab

Therefore,

x²-10x+25

=x²-2.x.5+5²

=(x-5)²

The error is in third step.

The correct solution is (x-5)²

The subtraction of square of b from square a is the product of sum of a and b and the subtraction a from b.

(a²-b²)=(a+b)(a-b)

16c²-9

=(4c)²-3²

=(4c+3)(4c-c)

The error is in third step.

The correct solution is (4c+3)(4c-c)

for finding the midpoint of any given 2 pair of points simply add the respective x coordinates with other x coordinate and the y coordinate with other y coordinate and divide them by 2

in your question it will be :

(3+1)/2 for the x coordinate of mid point i.e 2

[8+(-4)]/2 for the y coordinate of midpoint i.e 2

so coordinates of mid point of points (3,8) and (1,-4) will be (2,2)

a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.