Answer:
t = 2.28 s
Step-by-step explanation:
h = 105 - 9t - 16t ^ 2
0 ft = 105 ft - 9t -16^t
To find the roots of a quadratic function we have to use the Bhaskara formula
, the roots will give us the time it takes to reach zero height
ax^2 + bx + c = 0
-16^t - 9t + 105 ft = 0 ft
a = -16 b = -9 c = 105
t1 = (-b + √ b^2 - 4ac)/2a
t2 =(-b - √ b^2 - 4ac)/2a
t1 = (9 + √(-9^2 - (4 * (-16) * 105)))/2 * (-16)
t1 = (9 + √(-81 + 6720))/ -32
t1 = (9 + √6639)/ -32
t1 = (9 + 81.84)/ -32
t1 = 90.84 / -32
t1 = -2.83 s
t2 = (9 - √(-9^2 - (4 * (-16) * 105)))/2 * (-16)
t2 = (9 - √(-81 + 6720))/ -32
t2 = (9 - √6639)/ -32
t2 = (9 - 81.84)/ -32
t2 = -72.84 / -32
t2 = 2.28 s
we have two possible values, we are only going to take the positive one, beacause we are talking about time
t2 = 2.28 s
Answer: -0.22222222...
Step-by-step explanation:
-2 over 9 can be written as -2/9
So that’s basically -2 divided by 9. If you’d do long division properly, you should get -0.22222222...
1 1 9 . 6 5
1 is the the hundreds place
1 is the tens place
9 is the ones place
6 is the tenths place
5 is the hundredths place
Idkhzslkkxkfof?&$ask a another girl for that answer
<span>We have lateral faces that are rectangles and 2 congruent polygons that are bases. This must be 3-dimensional solid figure - quadrilateral prism. If those 2 bases are squares then it is a rectangular prism, but the bases also can be 2 rhombuses. Answer: Quadrilateral prism.</span><span />
