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°
Answer:
According to my calculations, 1.6 goes into 10.8 a total of 6 times with a remainder of 1.1999999999999993. If you continue the long division beyond the decimal point, the result would be 6.75.
Step-by-step explanation:
probability that a dessert sold at a certain cafe contains chocolate is 86%.
The probability that a dessert containing chocolate also contains nuts is 30%.
Find the probability that a dessert chosen at random contains nuts given that it contains chocolate
P(nuts given chocolate) = .30/.86 = .349 or 34
Answer:
A
Step-by-step explanation:
v=πr2h
r=(3)²* 5
45π unit³
