Answer:
the first and last one apply
Step-by-step explanation:
just did it on edu
-3(y+2)^2-5+6y
-3(y^2+4y+4)-5+6y
-3y^2-12y-12-5+6y
-3y^2-6y-17
∠ ABD = 5(2X+1)
∠ DBC = 3X+6
∠ EBC = Y +135/2
∠ ABD and ∠ DBC are linear pairs
∴ ∠ ABD +∠ DBC = 180
∴ 5(2X+1) + 3X+6 =180
solve for x
∴ x = 13
∴∠ ABD = 5(2X+1) = 5(2*13+1) = 135
∠ DBC = 3x+6 = 3*13+6 = 45
∠ ABD and ∠ EBC are vertical angles
∴ ∠ ABD = ∠ EBC = 135
∴ y +135/2 = 135
∴ y = 135/2
The <span>statements that are true:
--------------------------------------</span><span>
C.) x=13
E.)measure of angle EBC =135
F.) angle DBC and angle EBC are linear pairs
</span>
Answer:
see below
Step-by-step explanation:
1.5x + 5y = 1152
x = 4y – 2
We can substitute the second equation into the first equation
Which one-variable linear equation can be formed using the substitution method?
1.5(4y-2) +5y = 1152
Distribute
6y -3 +5y = 1152
Combine like terms
11y-3 = 1152
Add 3 to each side
11y-3+3 = 1152+3
11y = 1155
Divide each side by 11
11y/11 = 1155/11
y = 105
How many $5 raffle tickets were sold?
105 5 dollar tickets were sold
Now we need to find the number of 1.50 tickets
Which equation can be used to determine how many $1.50 raffle tickets were sold?
x = 4y – 2
x = 4(105) -2
=420-2
= 418
How many $1.50 raffle tickets were sold?
418 $1.50 tickets were sold
