Answer:
4x-y+1=0
Step-by-step explanation:
here,given equation of a line id
4x-y-2=0.. eqn(i)
equation of any line parallel to line (i) is
4x-y+k=0...eqn(ii)
since, the line(ii) passes through (1,5)[replacing x=1 and y=5 in eqn(ii), we get]
4*1-5+k=0
or, 4-5+k=0
or,-1+k=0
•°•k=1
substituting the value of k=1 in eqn(ii),
4x-y+1=0 is the required equation of the line.
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 multiply 18 by 8 and you get 144
so the puppys weight is 144oz now
Answer:
Step-by-step explanation:
Given
-- degree of freedom
Required
Determine the t-value
The given parameters can be illustrated as follows:
Where
So, we have:
To solve further, we make use of the attached the student's t distribution table.
From the attached table,
The t-value is given at the row with df = 28 and is 3.673900
Hence,
