Download please solve these problems/show work

please solve these problems/show work
Divide
(12a squared - 25a -7 divided by (3a-7)
Solution. Since 12a 2  25a  7  (3a  7)(4a  1) , we have
(12a 2  25a  7)  (3a  7)  4a  1
(x cubed - 6x sqyared + 7x -2 divided by (x-1)
Solution. Since x 3  6 x 2  7 x  2  ( x  1)( x 2  5x  2) , we have
( x 3  6 x 2  7 x  2)  ( x  1)  x 2  5x  2
A circle has a radius of 10 inches. Find the increase in area that occurs when the radius is
increased by 2 in. Round to the nearest hundredth.
Solution. We know that r=10inches, r  2(inches ) . So, the increase in area is
A   (r  r ) 2  r 2   (12) 2   (10) 2  44 (inches 2 )  138.23(in 2 )
An object is released from the top of an building 320 ft high. The initial velocity is 16ft/s.
How many seconds later will the object hit the ground?
Solution. Assume that the acceleration of gravity is g  9.8(m / s 2 ) . Now we need to
convert feet to meters, we know that 1m=3.28(feet). So, the initial velocity is v0=16ft/s
is
4.878m/s. h=320ft=97.56(m). Use a formula
h  v0 t  12 gt 2
We have
97.56  4.878t  4.9t 2
So,
4.9t 2  4.878t  97.56  0
So,
 4*4.9*97.56
t   4.878 4.878
 194.62( s)
4.9*2
So, 194.62 seconds later the object will hit the ground
2
How many consecutive natural numbers beginning with 1 will give the sum of 120?
Solution. We know that 1  2  ...  n 
n ( n 1)
2
. So, if 1  2  ...  n 
n ( n 1)
2
 120 , then
n  n  240  0
2
So,
(n+16)(n-15)=0
So, n=15 (since n>0).
So, 15 consecutive natural numbers beginning with 1 will give the sum of 120
Write each fraction in terms of the LCM of the denominators
x-3 divided by 3x squared + 4x-4 ; 2 divided by x + 2
Solution.
x 3
3 x 2  4 x 4
and
2
x2
Since 3x 2  4 x  4  ( x  2)(3x  2) , the LCM of the denominators is 3 x 2  4 x  4 .
So,
x 3
= 3 x2x43x4
3 x 2  4 x 4
and
2
x2
2(3 x 2)

3 x 2  4 x 4
2x divided by 10 + 3x-x squared; x + 2 divided by x squared -8x + 15
Solution.
2x
 (5 x2)(x1 x )  ( x 32)(xx(x5)(3)x 1)
103 x  x 2
and

x2
x 2 8 x 15
x2
( x 3)( x 5)

( x  2 )( x 1)
( x 3)( x 5)( x 1)
Simplify
3 over 2x + 1 minus 3 divided by 2 minus 4 x over 2x + 1
Solution.
3
2 x 1
 2 34 x 
2 x 1
3
2 x 1
 2(32(x2x1)1)4 x 
3
2 x 1
 3( 22x1) 
63( 2 x 1) 2
2( 2 x 1)

312 x 12 x 2
2( 2 x 1)
2 plus 5 over x minus 12 over x squared divided by 4 minus 4 over x minus 3 over x
squared
Solution.
(2  5x  12
)  (4  4x  x32 )
x2

2 x 2  5 x 12
x2
 4x

2 x 2  5 x 12
x2
 4 x 2 x 4 x 3

2 x 2  5 x 12
4 x 2  4 x 3

2
 4 x 3
x2
2
x4
2 x 1
1+ 4 over x + 4 over x squared divided by 1- 2 over x - 8 over x squared
Solution.
(1  4x  x42 )  (1  2x  x82 )

x2 4 x4
x2

x2 4 x4
x 2  2 x 8

x2
x 4
x
2
 2 x 8
x2
1 minus 2x -2 over 3x -1 divided by x minus 4 over 3x -1
Solution.
(1  23xx12 )  ( x  3 x41 )
 3xx11  3x3 xx14  3 x14
2
x+3 minus 18 over 2x +1 divided by x minus 6 over 2x +1
Solution.
6
( x  3  218
x 1 )  ( x  2 x 1 )

2 x 2  7 x 15
2 x 1
 2 x2 xx16 
2
2 x 2  7 x 15
2 x 2  x 6

x 5
x2
Solve the formula for the variable given
C= 5 over 9 (F-32);F (temputure conversion)
Solution. Since C 
5
9 ( F 32 )
F  32 
, we have
5
9C
So,
F  32  95C 
288C 5
9C
T=fm-gm;m (engineering)
Solution. Since T  fm  gm , we have
m  f T g
a=S-Sr; S 9mathematics)
Solution. Since a  S  Sr , we have
S  1ar
Solve. round to the nearest hundreth
The two legs of a right triangle measures 8in and 4 in. Find the length of the hypotenuse.
Solution. The length of the hypotenuse is given by
8 2  4 2  80  8.94(in )
Marta leaves a dock in her sailboat and sails 2.5 mi due east. She then tacks and sails 4
mi due north. The walkie talkie Marta has on board has a range of 5 mi. Will she be able
to call a friend on the dock from her location using the walkie talkie?
We can compute the distance from the beginning to the end, namely, AC. We have
AC  AB 2  BC 2  2.5 2  4 2  4.72(mi)  5(mi)
So, she will be able to call a friend on the dock from her location using the walkie talkie.
How far would a submarine periscope have to be above the water for the lookout to
locate a ship 7 mi away? The equation for the distance in miles that the lookout can see is
d= square root 1.5 h, where h is the height in feet above the surface of the water.
Solution. Since d  1.5h , we have
h
d
1.5

7
1.5
 5.72(mi)
The speed of a child riding a merry go round is givne by the equation v= square root 12r,
where v is the speed in feet per second and r is the distance in feet from the center of the
merry go round to the rider. If a child is moving 15 ft/s, how far is the child from the
center of the merry go round?
Solution. Since v  12r , we have
r
v
12

15
12
 4.33( ft)
Find the length of a pendulum that makes one swing in 2s. The equation for the time of
one swing is T=2 pi square root L over 32, where T is the time in seconds and L is the
length in feet.
Solution. Since T  2 L / 32 , we have
L  32( 2T ) 2  32( 22 ) 2  3.24( feet )
A commuter plane leaves an airport traveling due south at 400mph. Another plane
leaving at the same time travels due east at 300mph. Find the distance between the two
planes after 2 h.
Solution. See a graph.
We have
AB=400*2=800(m)
AC=300*2=600(m)
So,
BC  AB 2  AC 2  800 2  600 2  1000(m)
So, the distance between the two planes after 2 h is 1000m.
A stone is dropped into a mine shaft and hits the bottom 3.5 s later. How deep is the
mione shaft? The equation for the distance an object falls in T seconds is T=square root d
over 16, where d is the distance in feet.
Solution. Since T  d / 16 , we have
d  16T 2  16 * 3.5 2  196( ft)
A model rocket is launched with an initial velocity of 200 ft/s. The height, h, of the rocket
t seconds after launch is givne by h= -16t squared plus 200t. How many seconds after the
launch will the rocket be 300 ft above the ground. Round to the nearest hundreth.
Solution. Since h  16t 2  200t , let
300  16t 2  200t
We have
16t 2  200t  300  0
i.e.,
4t 2  50t  75  0
so,
t
50 502  4*4*75
4*2

50 1300
8
So,
t1 
50 502  4*4*75
4*2

50 1300
8
 1.75( s )
t 2  50 504*24*4*75  50 81300  10.75( s)
So, 1.75 seconds after the launch will the rocket be 300 ft above the ground.
2
In a slow pitch softball game, the height of the ball thrown can be modeled by the
equation h= -16t squared plus 24t plus 4, where h is the height of the ball in feet, and t is
the time, in seconds, since it was released by the pitcher. If the batter hits the ball when it
is 2 ft off the ground, how many seconds has the ball been in the air? Round to the
nearest hundreth.
Solution. Since h  16t 2  24t  4 , by hypothesis, we know that h=2. So,
2  16t 2  24t  4
i.e.,
 16t 2  24t  2  0
i.e,
 8t 2  12t  1  0
So,
t
12 122 4*1*(8)
8*2
 12
14432
16
So,
t  12
NOTE: Since t>0
14432
16
 1.58(s)
The hang time of a football that is kicked is given by s= -16t squared plus 88t plu 1,
where s is the height. in feet, of the football t seconds after leaving the kicker's foot. What
is the hang time of a kickoff that hits the ground without being caught? Round to the
nearest tenth.
Solution. s  16t 2  88t  1
Since we don’t know the value of s, we couldn’t solve it to get t.