Let the score of cowboys is x
and giants make score 9 which is twice less than the cowboys score so
giants score will be = 2x -9
and packers scored 14 more than giants that is (2x - 9) + 14
now sum of their scores is equal to 81 it means:
x + (2x - 9) + (2x -9) + 14 = 81
x + 2x - 9 + 2x - 9 + 14 = 81
5x = 81 + 4
5x = 85
x = 17
packers scored = (2x - 9) + 14
= 2 (17) -9 + 14
=38 + 5 = 43 points
Answer:
Y= 10x-24 would be parralel to y=10x+6
Step-by-step explanation:
well a parralel line has to have the same slope and a different y-intercept.
To find the y-intercept the x has to be exactly 0.
So we have (2,-4)
The slope is 10/1 which is 10y and 1 x which in an ordered park would be Add 1 x and add 10 y. But we’re trying to get x to be 0 to find the y-intercept.
So subtract 1 x and subtract 10y
2-1=1
-4-10=-14
(1, -14)
1-1=0
-14-10=-24
(0, -24)
THe y-intercept is -24 and w have the slope since its the same thing So
y=10x-24 is the equation
Answer:
if im right 1600? i did it in my head so im not 100 percent sure
Step-by-step explanation:
The required proof is given in the table below:
![\begin{tabular}{|p{4cm}|p{6cm}|} Statement & Reason \\ [1ex] 1. $\overline{BD}$ bisects $\angle ABC$ & 1. Given \\ 2. \angle DBC\cong\angle ABD & 2. De(finition of angle bisector \\ 3. $\overline{AE}$||$\overline{BD}$ & 3. Given \\ 4. \angle AEB\cong\angle DBC & 4. Corresponding angles \\ 5. \angle AEB\cong\angle ABD & 5. Transitive property of equality \\ 6. \angle ABD\cong\angle BAE & 6. Alternate angles \end{tabular}](https://tex.z-dn.net/?f=%20%5Cbegin%7Btabular%7D%7B%7Cp%7B4cm%7D%7Cp%7B6cm%7D%7C%7D%20%0A%20Statement%20%26%20Reason%20%5C%5C%20%5B1ex%5D%20%0A1.%20%24%5Coverline%7BBD%7D%24%20bisects%20%24%5Cangle%20ABC%24%20%26%201.%20Given%20%5C%5C%0A2.%20%5Cangle%20DBC%5Ccong%5Cangle%20ABD%20%26%202.%20De%28finition%20of%20angle%20bisector%20%5C%5C%20%0A3.%20%24%5Coverline%7BAE%7D%24%7C%7C%24%5Coverline%7BBD%7D%24%20%26%203.%20Given%20%5C%5C%20%0A4.%20%5Cangle%20AEB%5Ccong%5Cangle%20DBC%20%26%204.%20Corresponding%20angles%20%5C%5C%0A5.%20%5Cangle%20AEB%5Ccong%5Cangle%20ABD%20%26%205.%20Transitive%20property%20of%20equality%20%5C%5C%20%0A6.%20%5Cangle%20ABD%5Ccong%5Cangle%20BAE%20%26%206.%20Alternate%20angles%0A%5Cend%7Btabular%7D)
2 right 3 up... hope this helped you