i think the answer is c 1.4
Answer:
(0.88, 0,65)
Step-by-step explanation:
what do we have to do ? I assume we need to find the crossing point between the 2 lines.
when looking at the graphic, we can already rule out the first, third and fourth option.
none of the coordinates of that point is larger than 1 in any direction or smaller than 0.5.
so, just by looking at things I immediately "bet" on answer option 2 (0.88, 0.65).
but let's check.
y = -2x/5 + 1
y = 3x - 2
for the crossing point both functions must produce the same functional value.
=>
-2x/5 + 1 = 3x - 2
-2x/5 + 5/5 = 15x/5 - 10/5
-2x + 5 = 15x - 10
15 = 17x
x = 15/17 = 0.88...
y = 3×0.88... -2 = 0.65...
hurray, our original observation and "suspicion" is confirmed.
Answer:
<em>option: C </em>is correct.
Step-by-step explanation:
since, we are given a parent function f(x) as: f(x)=x.
We are asked to find what change will occur in the function when the parent function f(x) is replaced by f(9x).
let g(x) be the function formed by replacing f(x) with f(9x) i.e.
g(x)=f(9x)=9x.
clearly the slope of the function g(x)=f(9x) is 9. ( since on comparing the equation with the slope intercept form <em>y=mx+c</em>; where <em>m</em> is the<em> slope</em> of the line and <em>c</em> is the <em>y- intercept</em><em>)</em>.
whereas the slope of the function f(x)=x is 1.
<em>Hence there is an increment of slope by a factor of 9.</em>
<em>Hence, option C is correct.</em>
Answer: I think A
Step-by-step explanation:
Answer:
![t=1.7 \ sec\ , t=4.0\ sec](https://tex.z-dn.net/?f=t%3D1.7%20%5C%20sec%5C%20%2C%20t%3D4.0%5C%20sec)
Step-by-step explanation:
<u>Vertical Throw</u>
It refers to a situation where an object is thrown verticaly upwards with some inicial speed v_o and let in free air (no friction) until it completes its movement up and finally returns to the very same point of lauch. The only acting force is gravity
The projectile formula is given as
![h=-16t^2+v_0t](https://tex.z-dn.net/?f=h%3D-16t%5E2%2Bv_0t)
where t is time in seconds, h is the height in feet and v is the speed in ft/sec
We are required to find the time t where h=120 ft, knowing ![v_o=90\ ft/sec](https://tex.z-dn.net/?f=v_o%3D90%5C%20ft%2Fsec)
![-16t^2+90t=120](https://tex.z-dn.net/?f=-16t%5E2%2B90t%3D120)
Rearranging
![-16t^2+90t-120=0](https://tex.z-dn.net/?f=-16t%5E2%2B90t-120%3D0)
This is a second-degree equation which will be solved with the formula
![\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
![\displaystyle x=\frac{-90\pm \sqrt{90^2-4(-16)(-120)}}{2(-16)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B-90%5Cpm%20%5Csqrt%7B90%5E2-4%28-16%29%28-120%29%7D%7D%7B2%28-16%29%7D)
![\displaystyle x=\frac{-90\pm \sqrt{1380}}{-32}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B-90%5Cpm%20%5Csqrt%7B1380%7D%7D%7B-32%7D)
Two solutions are obtained
![\boxed{t=1.7 \ sec\ , t= 4.\ sec}](https://tex.z-dn.net/?f=%5Cboxed%7Bt%3D1.7%20%5C%20sec%5C%20%2C%20t%3D%204.%5C%20sec%7D)
Both solutions are possible because the ball actually is at 120 ft in its way up and then when going down