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°
XY=10
Y=X+3
X(X+3)=10
X^2+3X=10
X^2+3X-10
(X-2)(X+5)
Answer:
x = 10
Step-by-step explanation:
Using the rules of logarithms
• log x + log y ⇔ log(xy)
• log
⇔ n log x
• log x = log y ⇒ x = y
Given
ln20 + ln5 = 2 lnx , then
ln(20 × 5) = ln
ln 100 = ln, hence
x² = 100 ( take the square root of both sides )
x = 10
Replace the variables with numbers to find the expression
domain of the problem is -5 and the range is 4
