She could actually miss and get a 0 because 108+72+95/3=91
1 Remove parentheses.
2(5c−7)≥c−3
2 Expand.
10c−14 ≥ c −3
3 Add 14 to both sides.
10c-14+14≥ c − 3+14
4 Simplify c−3+14 to c+11
10c≥c+11
5 Subtract c from both sides.
10c−c≥11
6 Simplify 10c−c to 9c
9c≥11
7 Divide both sides by 9.
c ≥11/9
Well, We have to Know What The Distance is to the Finish Line. product of is multiplication.
The answer is very simple .
First you need to determine the slope, which is the reciprocal of the first line.
Then,
y = mx + b
m => -1/m
The perperdicular line has a new slope equal to m = -1/2
In the formula (y - y1) = m ( x - x1 ) + b , you should sustitude this value.
So the final equation will be
-3 = -1/2 (0) + b
b = - 3
y = (-1/2)x - 3
That is the solution
Answer:
Q1. x= 18, y=59
Q2. m∠J= 56°
Step-by-step explanation:
Q1. (3x +5)°= y° (base ∠s of isos. △)
y= 3x +5 -----(1)
(3x +5)° +y° +(4x -10)°= 180° (∠ sum of △)
3x +5 +y +4x -10= 180
7x +y -5= 180
7x +y= 180 +5
7x +y= 185 -----(2)
Substitute (1) into (2):
7x +3x +5= 185
10x= 185 -5
10x= 180
x= 180 ÷10
x= 18
Substitute x= 18 into (1):
y= 3(18) +5
y= 59
Q2. (5x -13)°= (3x +17)° (base ∠s of isos. △)
5x -13= 3x +17
5x -3x= 17 +13
2x= 30
x= 30 ÷2
x= 15
∠LKJ
= 3(15) +17
= 62°
∠KLJ= 62° (base ∠s of isos. △)
m∠J
= 180° -62° -62° (∠ sum of △JKL)
= 56°
