H(t) = -16t² + 60t + 95
g(t) = 20 + 38.7t
h(1) = -16(1²) + 60(1) + 95 = -16 + 60 + 95 = -16 + 155 = 139
h(2) = -16(2²) + 60(2) + 95 = -16(4) + 120 + 95 = -64 + 215 = 151
h(3) = -16(3²) + 60(3) + 95 = -16(9) + 180 + 95 = -144 + 275 = 131
h(4) = -16(4²) + 60(4) + 95 = -16(16) + 240 + 95 = -256 + 335 = 79
g(1) = 20 + 38.7(1) = 20 + 38.7 = 58.7
g(2) = 20 + 38.7(2) = 20 + 77.4 = 97.4
g(3) = 20 + 38.7(3) = 20 + 116.1 = 136.1
g(4) = 20 + 38.7(4) = 20 + 154.8 = 174.8
Between 2 and 3 seconds.
The range of the 1st object is 151 to 131.
The range of the 2nd object is 97.4 to 136.1
h(t) = g(t) ⇒ 131 = 131
It means that the point where the 2 objects are equal is the point where the 1st object is falling down while the 2nd object is still going up.
Answer:x=31.4919
Step-by-step explanation:
Step1:isolate a square root on the left hand side√x+3=√2x-1-2
Step2:eliminate the radicals on the left hand side
Raise both sides to the second power
√x+3)^2=(√2x-1-2)^2
After squaring
x+3=2x-1+4-4-4√2x-1
Step3:get the remaining radicals by itself
x+3=2x-1+4-4√2x-1
Isolate radical on the left hand side
4√2x-1=-x-3+2x-1+4
4√2x-1=x
Step4:eliminate the radicals on the left hand side
Raise both side to the second power
(4√2x-1)^2=x^2
After squaring
32x-16=x^2
Step 5:solve the quadratic equation
x^2-32x-16
This equation has two real roots
x1=32+√960/2=31.4919
x2=32-√960/2=0.5081
Step6:check that the first solution is correct
Put in 31.4919 for x
√31.4919+3=√2•31.4919-1-2
√34.492=5.873
x=31.4919
Step7:check that the second solution is correct
√x+3=√2x-1-2
Put in 0.5081 for x
√0.5081+3=√2•0.5081-1-2
√3.508=-1.873
1.873#-1.873
One solution was found
x=31.4919
The answer would be 20,670
1+4=5
2+5=7
5+7=12
3+6=9
12+9=21
I do not see21 up here so please cheek your question again or please change the 19 to a 21
