Input the equation for the following line in standard form Ax + By =c. Reduce all fractional answers to lowest terms. A line with y-intercept of 2 that is parallel to 2x + y = -5.
Given two points on the line, input the equation of the line in standard form Ax + By =
c. Reduce all fractional answers to lowest terms. (7, -3), (4, -8)
A line that passes through the origin and is parallel to x + y = 6.
Answers
Answer 1
Answer:
c. Reduce all fractional answers to lowest terms. A line with y-intercept of 2 that is parallel to 2x + y = -5
Related Questions
Hugh has to earn at least $400 to meet his fundraising goal. He has only 100 books that he plans to sell at $8 each. Which inequality shows the number of books, x, Hugh can sell to meet his goal?
A scale model of a building is 2.3 feet high. The ratio of the height of the model to the height of the building is 11:56. How tall is the building, in feet?
In the diagram below, the top of each stair is parallel to the floor. What is the value of y?(Picture below.)
What is the value of i^97 - i
Find what the measurements of angles a-c are and state the reasonPLEASE HELP
A scale model of a building is 2.3 feet high. The ratio of the height of the model to the height of the building is 11:56. How tall is the building, in feet?
In the diagram below, the top of each stair is parallel to the floor. What is the value of y?(Picture below.)
What is the value of i^97 - i
Find what the measurements of angles a-c are and state the reasonPLEASE HELP
Answers
The sum of solar energy can assumed as 'x', so it is easy to conclude that the sum of wind energy is '8x+56'. From the above assumption, we can easily get that 'x+8x+56=1037', to solve this equation and get tha
Answers
Answers
xy = -150
x + y = 5
x + y = 5
x - x + y = -x + 5
y = -x + 5
xy = -150
x(-x + 5) = -150
x(-x) + x(5) = -150
-x² + 5x = -150
-x² + 5x + 150 = 0
-1(x²) - 1(-5x) - 1(-150) = 0
-1(x² - 5x - 150) = 0
-1 -1
x² - 5x - 150 = 0
x = -(-5) ± √((-5)² - 4(1)(-150))
2(1)
x = 5 ± √(25 + 600)
2
x = 5 ± √(625)
2
x = 5 ± 25
2
x = 2.5 ± 12.5
x = 2.5 + 12.5 or x = 2.5 - 12.5
x = 15 or x = -10
x + y = 5
15 + y = 5
- 15 - 15
y = -10
(x, y) = (15, -10)
or
x + y = 5
-10 + y = 5
+ 10 + 10
y = 15
(x, y) = (-10, 15)
The two numbers that add up to 5 and multiply to -150 are 15 and -10.
x + y = 5
x + y = 5
x - x + y = -x + 5
y = -x + 5
xy = -150
x(-x + 5) = -150
x(-x) + x(5) = -150
-x² + 5x = -150
-x² + 5x + 150 = 0
-1(x²) - 1(-5x) - 1(-150) = 0
-1(x² - 5x - 150) = 0
-1 -1
x² - 5x - 150 = 0
x = -(-5) ± √((-5)² - 4(1)(-150))
2(1)
x = 5 ± √(25 + 600)
2
x = 5 ± √(625)
2
x = 5 ± 25
2
x = 2.5 ± 12.5
x = 2.5 + 12.5 or x = 2.5 - 12.5
x = 15 or x = -10
x + y = 5
15 + y = 5
- 15 - 15
y = -10
(x, y) = (15, -10)
or
x + y = 5
-10 + y = 5
+ 10 + 10
y = 15
(x, y) = (-10, 15)
The two numbers that add up to 5 and multiply to -150 are 15 and -10.
Answers
let the cost price of bench be x
then cost price of garden table is x+35
total cost = x + x+35
985 = 2x +35
985 -35÷2 =x
425 =×
then the total cost of bench is 425 and garden is 460
then cost price of garden table is x+35
total cost = x + x+35
985 = 2x +35
985 -35÷2 =x
425 =×
then the total cost of bench is 425 and garden is 460
985-35=950, 950÷2=475, 475+35=510, 510 and 475
Answers
A method to solve such problems is to consider the rate of change of the angle in degrees per minute. The hour hand of a normal 12-hour analogue clock turns 360° in 12 hours (720 minutes) or 0.5° per minute. The minute hand rotates through 360° in 60 minutes or 6° per minute.
Answers
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
Answer:
Step-by-step explanation:
362,880
Other Questions