comparing the relation given
thus y= -2x +1 to the general equation of a line
where y= mx + c
where m is the gradient and c is the intercept on the y-axis.
from the question, the gradient is -2 and since the line is perpendicular, the gradient is given as
so substituting the value of m in to the equation
=
as the gradient
equation of line is given by
y - y1 = m( x-x1)
from the question y1= -2 and x1=1
substitute them
y-(-2) = 1/2 (x -1)
y+2 =1/2 (x-1)
multiplying through by 2
2 (y+2) = x-1
2y +4 = x-1
2y =x-1-4
2y =x-5
x -2y -5 = 0
therefore the equation for the line is x-2y -5=0
Answer:
The answer is (D) the starting temperature was 82 degrees F and dropped at a rate of 8 degrees per hour.
Step-by-step explanation:
For every 2 hours the temperature drops 8 degrees.
hope it helps
Answer: C
please mark brainliest
Answer:
D=24
Step-by-step explanation:
y inversely as x
x=k/y
By cross multiplication
k=xy
k=12*6
k=72
finding y
72=3y
y=24
Answer:
Answer for 2nd is option c, for 3rd is option d, for 4th is option e
Step-by-step explanation:
As we know 1 ft.=12 in.
- In ΔABC
∴ The congruent sides are AB and AC respectively
- CB =12 ft. 4 in.=148 in.
- AB=
CB =111 in. =9 ft. 3 in. - AC=CB =111 in. =9 ft. 3 in.
∵ <em>Perimeter of ΔABC</em> =AB+AC+CB
=9 ft. 3 in. + 9 ft. 3 in. +12 ft. 4 in.
=30 ft. 10 in.
2. In ΔDEF
∴ The congruent sides are DE and DF respectively
- DE = 6 ft. 3 in. =75 in.
- DF = 6 ft. 3 in. =75 in.
- Let the length of FE is equal to x
- 0.75FE =DE =DF
- 0.75x = 6 ft. 3 in. =75 in.
- x =100 in. =8 ft. 4 in.
∵ <em>Perimeter of ΔDEF</em> =DE+DF+FE
= 6 ft. 3 in. +6 ft. 3 in. +8 ft. 4 in.
= 20 ft. 10 in.
3. In ΔJKL
∴ The congruent sides are JL and KL respectively
- JK = x+3
- KL =4x-17
- JL =6x-45
- JL≅KL
- 4x-17 =6x-45 . . . . . . . . . . . . . . . . . . . . . . . (1)
- Subracting 4x from both sides from eq 1
- -17 =2x-45
- Adding 45 on both the sides
- 28 =2x
- Dividing by 2 on both sides
- 14 =x
- JK = 14+3 =17
- KL = 4×14-17 =39
- JL = 6×14-45 =39
∵ <em>The dimensions of the ΔJKL are 39,39 and 17.</em>
