Answer:
m∠DEF = 50
Step-by-step explanation:
∠DEG = ∠GEF
3y + 4 = 5y - 10
subtract 4 from both sides
3y = 5y - 14
subtract 5y from both sides
-2y = -14
divide by -2
y = 7
Add 7 into the equations:
3y + 4 + 5y - 10
3(7) + 4 + 5(7) - 10
21 + 4 + 35 - 10
25 + 25
50
Answer:
The answer to your question is: h = 14.5 in
Step-by-step explanation:
Data
radius = 5 in
Volume = 725 π in³
height = ?
Formula
Volume = 2πr²h
Substitution
725 π in³ = 2πr² h
h = 725π / 2πr²
h = 725 / 2(5)²
h = 725 / 50
h = 14.5 in
Answer:
Step-by-step explanation:a or b ?
The values are x=8 and y=35, if the given ΔABC and ΔDEC are equal, it is obtained by Pythagoras theorem.
Step-by-step explanation:
The given are,
From ΔABC,
AB= 6
BC= 10
AC = x
From ΔDEC,
CD= 28
DE= 21
CE = y
Step:1
Pythagoras theorem from ΔABC,
...............(1)
Substitute the values,
= 
+ 
100 = 36 +
= 100 - 36
= 64
AC = 
AC = 8
AC = x = 8
Step:2
Pythagoras theorem for ΔDEC,
................(2)
From the values,
= 
+ 
= 784 + 441
= 1225
CE = 
CE = 35
CE = y = 35
Result:
The values are x=8 and y=35, if the given ΔABC and ΔDEC are equal.
Answer:
Time = 3secs
Max height = 56m
Step-by-step explanation:
Given the height reached by the rocket modeled by the equation;
h(x) = -5(x - 3)² + 56 where;
x is in seconds
The rocket velocity at its maximum height is zero.
Hence dh/dx = 0
dh/dx = 2(-5)(x-3)
dh/dx = -10(x-3)
Since dh/dx = 0
0 = -10(x-3)
0 = -10x + 30
10x = 30
x = 3secs
<em>Hence it takes 3 secs to reach the maximum height.</em>
Get the maximum height reached by the rocket. Substitute x = 3 into the equation given;
Recall that:
h(x) = -5(x - 3)² + 56
If x = 3
h(3) = -5(3 - 3)² + 56
h(3) = 0 + 56
h(3) = 56
<em>Hence the maximum height that the rocket reaches is 56m</em>
