When the height is 0 that is when it hits the ground
solve
0=-16t²+25x+3
cant factor so use quadratic formula
for
0=ax²+bx+c
so for
0=-16t²+25t+3
a=-16
b=25
c=3
so
it has to be a positive time
so it will hit the ground after
seconds
<span>For given hyperbola:
center: (0,0)
a=7 (distance from center to vertices)
a^2=49
c=9 (distance from center to vertices)
c^2=81
c^2=a^2+b^2
b^2=c^2-a^2=81-49=32
Equation of given hyperbola:
..
2: vertices (0,+/-3) foci (0,+/-6)
hyperbola has a vertical transverse axis
Its standard form of equation: , (h,k)=(x,y) coordinates of center
For given hyperbola:
center: (0,0)
a=3 (distance from center to vertices)
a^2=9
c=6 (distance from center to vertices)
c^2=36 a^2+b^2
b^2=c^2-a^2=36-9=25
Equation of given hyperbola:
</span>
A. Is the attached image. The slope is 7.5x
b. 1 = 7.5x
x = 1.33333…
This is hours so we must multiply this by 60
x = 8 min
Answer:10 people
Step-by-step explanation:
7 times less showed up so, 70 divided by 7 is 10. 10 guests attended.
Answer:
m = 10
Step-by-step explanation:
The value of <em>m</em> that would make this equation true is <em>10</em>. To figure this out you must work the equation to combine like terms. To start, remember PEMDAS. You would begin with <em>1/2 (8m - 18) </em>and multiply both <em>8m </em>and <em>18 </em>by <em>1/2. </em>Because half of <em>8</em> is <em>4</em> and half of<em> 18</em> is <em>9</em>, your new equation would be <em>4m - 9 = 31. </em>From here you would add nine to both sides to finish combining like terms. The equation from this point should be <em>4m = 40.</em> To find the value of <em>m</em>, you then have to divide both sides by <em>4</em>, leading to the equation/solution of <em>m = 10.</em>
