Parallel lines have the same slope, but different y-intercepts
The slope of both lines is -2/3
To find the equation that passes through (7, 3) we can use the point-slope formula: y - y1 = m (x - x1)
y - (3) = (-2/3) (x - (7))
y - 3 = -2/3x + 4 and 2/3
y = -2/3x + 7 and 2/3
or
y = -2/3x + 7.67
:)))
V = 1/3 l h w
V = 1/3 (12)(15)(9)
V= = 540
answer 540 m^3
Part 1: Answer:
(x+1)(x+1)(x-6) = x^3 - 4x^2 - 11x - 6
Step-by-step explanation:
To make r a root, include (x-r) as a factor. (-1+1)(-1+1)(-1-6) is zero even though (-1-6) isn't.
(6+1)(6+1)(6-6) is zero.
Part 2 Answer:
Standard form: y = -x^4 + 12
Degree 4
left end goes down, right end goes down.
Step by step: apply the definitions of standard form, polynomial degree, and "end behavior". In other words, read the textbook.
Part 3: Answer: x = 3, x = 8
Step by step:
x^2-11x = -24
x^2-11x+24 = 0
(x-3)(x-8) = 0
x = 3 or x = 8
Part 4a Answer:
quotient 2x^2 + x - 3
remainder 1
Step by step:
2x^2 + x - 3
___________________
x-4 ) 2x^3 - 7x^2 - 7x + 13
2x^3 - 8x^2
__________
0 + x^2 - 7x + 13
x^2 - 4x
____________
0 - 3x + 13
- 3x + 12
______
1
Part 4b answer:
quotient 2x^2 - 6x + 2
remainder -20
Step by step: you have to know exactly what you are doing. Refer to textbook or Wikipedia.
dividend 2x^3 +14x^2 - 58x
divisor x+10
leading coefficient of divisor must be 1
write coefficients of dividend at top
write coefficients of dividend at left
| 2 14 -58 0
-10 | -20 60 -20
___________
| 2 -6 2 -20
Coefficients of quotient are 2 -6 2
Remainder is -20
quotient = 2x^2 - 6x + 2
An acute angle is infact 90 degrees, therefore your answer is correct.
