Answer: 144 pounds and 3 cents to the fourth.
Step-by-step explanation: some guy
Answer:
<h2>For c = 5 → two solutions</h2><h2>For c = -10 → no solutions</h2>
Step-by-step explanation:
We know

for any real value of <em>a</em>.
|a| = b > 0 - <em>two solutions: </em>a = b or a = -b
|a| = 0 - <em>one solution: a = 0</em>
|a| = b < 0 - <em>no solution</em>
<em />
|x + 6| - 4 = c
for c = 5:
|x + 6| - 4 = 5 <em>add 4 to both sides</em>
|x + 6| = 9 > 0 <em>TWO SOLUTIONS</em>
for c = -10
|x + 6| - 4 = -10 <em>add 4 to both sides</em>
|x + 6| = -6 < 0 <em>NO SOLUTIONS</em>
<em></em>
Calculate the solutions for c = 5:
|x + 6| = 9 ⇔ x + 6 = 9 or x + 6 = -9 <em>subtract 6 from both sides</em>
x = 3 or x = -15
So y+x=2
x-y=4
one way is subsituteion where yousubisutte one of the number for another exg
x-y=4
add y to both sides
x=y+4
subsitute y+4 for x in x+y=2
y+4+y=2
2y+4=2
subtract 4 from both sides
2y=-2
divide both sides by 2
y=-1
subsiutte
x+y=2
x-1=2
x=3
or elimination
x+y=2
x-y=4
add the two equations and cancel the y terms
x+x+y-y=2+4
2x=6
divide both sides by 2
x==3
subsitute
x+y=2
3+y=2
y=-1
so x=3
y=-1
Answer:

Step-by-step explanation:
<h2><u>IdEnTiTy UsED</u></h2>

1)

=(x-5)²
2)



3)

= (1+x)²
For this case we have the following equation:
h = -16t2 + 32t + 6
Deriving we have:
h '= -32t + 32
Equaling to zero:
-32t + 32 = 0
We clear the time:
t = 32/32
t = 1 s
We are now looking for the maximum height:
h (1) = -16 * (1) ^ 2 + 32 * (1) + 6
h (1) = 22 feet
Answer:
The ball reaches its maximum height in:
t = 1 s
The ball's maximum height is:
h (1) = 22 feet
option: 22 ft, 1 s