Answer:
the actual pressure is 31.09
Complete question:
What is the actual pressure when Ali's gauge shows 33.58
Step-by-step explanation:
Given,
Ali's tyre presser gauge show a reading 8% higher than the actual pressure.
Ali's gauge shows 33.58.
Let the actual pressure be x.
According to problem,
{ [If 8% increasing of a no ⇒ The number becomes =
of the number}
x= 
x=31.09
Therefore, the actual pressure is 31.09
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°
You would find the sign if the signs are not the same if you are dividing or multiplying its negitve if its the same its positive
Answer:
do yourself a favor and download photomath. It'll save your life
Step-by-step explanation: