Answer:(2x - 5)(3x + 1)
Step-by-step explanation:
6x^2 - 13x - 5
6x^2 + 2x - 15x - 5
2x (3x + 1) - 5 (3x + 1)
(2x - 5)(3x + 1)
Answer:
b
Step-by-step explanation:
In general
Given
y = f(x) then y = f(Cx) is a horizontal stretch/ compression in the x- direction
• If C > 1 then compression
• If 0 < C < 1 then stretch
Consider corresponding points on the 2 graphs
(2, 2 ) → (4, 2 )
(4, - 2 ) → (8, - 2 )
Indicating a stretch in the x- direction.
y = f(
) with C =
, that is 0 < C < 1
stretches the graph in the x- direction by a factor of 2
Thus
y = f(
) → b
Answers:
Angle A = 104 degrees
Angle B = 76 degrees
=================================
Supplementary angles add to 180 degrees.
A+B = 180
(3x-7)+(2x+2) = 180
5x-5 = 180
5x-5+5 = 180+5 ... add 5 to both sides
5x = 185
5x/5 = 185/5 .... divide both sides by 5
x = 37
Use x =37 to find
Angle A = 3x-7 = 3*37-7 = 111-7 = 104 degrees
Angle B = 2x+2 = 2*37+2 = 74+2 = 76 degrees
As a check: A+B = 104+76 = 180 so that helps confirm the answer
Answer:
The buses will cross after 3 hours at a distance of 135 km away from Anandpur
Step-by-step explanation:
Distance between Anandpur and Shivnagar 255 km
Bus 1 leave at 1:00 p.m. Its speed is 45 km
Bus 2 leaves at the same time with a speed of 40 km/hour.
Now the Let the time at which ther across each other be x . then
Distance covered Bus 1 at t time +distance covered by Bus 2 at t time = distance between anandpur and shivnagar-----------------(1)
<u>Distance covered Bus 1 at t time</u>
Distance = speed x time
Distance = 45 t
<u>Distance covered by Bus 2 at t time</u>
Distance = speed x time
Distance = 40 t
Now, Substituting then values in (1)
45t +40t = 255
85t = 255
t =![\frac{255}{85}](https://tex.z-dn.net/?f=%5Cfrac%7B255%7D%7B85%7D)
t = 3 hour
anandpur speed to reach at 4:00PM
distance =time×speed
d=3×45
135 km away from anandpur
<span> x - 4y = -5
-x + 6y = 7
---------------add
2y = 2
y = 1
</span> x - 4y = -5
x - 4(1) = -5
x - 4 = -5
x = -1
answer
x = -1 and y = 1