Try It
row: x x | y= 1 2 x+2 y= 1 2 x+2 | ( x,y ) ( x,y )
row: −2 −2 | y= 1 2 ( −2 )+2=1 y= 1 2 ( −2 )+2=1 | ( −2,1 ) ( −2,1 )
row: −1 −1 | y= 1 2 ( −1 )+2= 3 2 y= 1 2 ( −1 )+2= 3 2 | ( −1, 3 2 ) ( −1, 3 2 )
row: 0 0 | y= 1 2 ( 0 )+2=2 y= 1 2 ( 0 )+2=2 | ( 0,2 ) ( 0,2 )
row: 1 1 | y= 1 2 ( 1 )+2= 5 2 y= 1 2 ( 1 )+2= 5 2 | ( 1, 5 2 ) ( 1, 5 2 )
row: 2 2 | y= 1 2 ( 2 )+2=3 y= 1 2 ( 2 )+2=3 | ( 2,3 ) ( 2,3 )
x-intercept is ( 4,0 ); ( 4,0 ); y-intercept is ( 0,3 ). ( 0,3 ).
125 =5 5 125 =5 5
( −5, 5 2 ) ( −5, 5 2 )
x=−5 x=−5
x=−3 x=−3
x= 10 3 x= 10 3
x=1 x=1
x=− 7 17 . x=− 7 17 . Excluded values are x=− 1 2 x=− 1 2 and x=− 1 3 . x=− 1 3 .
x= 1 3 x= 1 3
m=− 2 3 m=− 2 3
y=4x−3 y=4x−3
x+3y=2 x+3y=2
Horizontal line: y=2 y=2
Parallel lines: equations are written in slope-intercept form.
y=5x+3 y=5x+3
11 and 25
C=2.5x+3,650 C=2.5x+3,650
45 mi/h
L=37 L=37 cm, W=18 W=18 cm
250 ft2
−24 =0+2i 6 −24 =0+2i 6
(3−4i)−(2+5i)=1−9i (3−4i)−(2+5i)=1−9i
5 2 −i 5 2 −i
18+i 18+i
−3−4i −3−4i
−1 −1
( x−6 )( x+1 )=0;x=6, x=−1 ( x−6 )( x+1 )=0;x=6, x=−1
( x−7 )( x+3 )=0, ( x−7 )( x+3 )=0, x=7, x=7, x=−3. x=−3.
( x+5 )( x−5 )=0, ( x+5 )( x−5 )=0, x=−5, x=−5, x=5. x=5.
( 3x+2 )( 4x+1 )=0, ( 3x+2 )( 4x+1 )=0, x=− 2 3 , x=− 2 3 , x=− 1 4 x=− 1 4
x=0,x=−10,x=−1 x=0,x=−10,x=−1
x=4± 5 x=4± 5
x=3± 22 x=3± 22
x=− 2 3 , x=− 2 3 , x= 1 3 x= 1 3
5 5 units
1 4 1 4
25 25
{ −1 } { −1 }
0, 0, 1 2 , 1 2 , − 1 2 − 1 2
1; 1; extraneous solution − 2 9 − 2 9
−2; −2; extraneous solution −1 −1
−੧, −੧, ੩ ੨ ੩ ੨
−੩,੩,−ੲ,ੲ −੩,੩,−ੲ,ੲ
੨,੧੨ ੨,੧੨
−੧ −੧ (੦੦ ਕੋਈ ਹੱਲ ਨਹੀਂ ਹੈ)।
[ −੩,੫ ] [ −੩,੫ ]
( −∞,−੨ )∪[ ੩,∞ ) ( −∞,−੨ )∪[ ੩,∞ )
x<੧ x<੧
x≥−੫ x≥−੫
( ੨,∞ ) ( ੨,∞ )
[ − ੩ ੧੪ ,∞ ) [ − ੩ ੧੪ ,∞ )
੬<x≤੯ ਜਾਂ ( ੬,੯ ] ੬<x≤੯ ਜਾਂ ( ੬,੯ ]
( − ੧ ੮ , ੧ ੨ ) ( − ੧ ੮ , ੧ ੨ )
| x−੨ |≤੩ | x−੨ |≤੩
k≤੧ k≤੧ ਜਾਂ k≥੭; k≥੭; ਅੰਤਰਾਲ ਨੋਟੇਸ਼ਨ ਵਿੱਚ, ਇਹ (−∞,੧]∪[੭,∞) ਹੋਵੇਗਾ। (−∞,੧]∪[੭,∞)।
੨.੧ ਭਾਗ ਅਭਿਆਸ
ਜਵਾਬ ਵੱਖੋ-ਵੱਖਰੇ ਹੋ ਸਕਦੇ ਹਨ। ਹਾਂ। ਇਹ ਸੰਭਵ ਹੈ ਕਿ ਇੱਕ ਬਿੰਦੂ x-ਧੁਰੇ ਜਾਂ y-ਧੁਰੇ 'ਤੇ ਹੋਵੇ ਅਤੇ ਇਸ ਲਈ ਇਸਨੂੰ ਕਿਸੇ ਵੀ ਚਤੁਰਭੁਜ ਵਿੱਚ ਨਾ ਮੰਨਿਆ ਜਾਵੇ।
y-ਅੰਤਰਖੰਡ ਉਹ ਬਿੰਦੂ ਹੈ ਜਿੱਥੇ ਗ੍ਰਾਫ y-ਧੁਰੇ ਨੂੰ ਕੱਟਦਾ ਹੈ।
x-ਅੰਤਰਖੰਡ ( ੨,੦ ) ( ੨,੦ ) ਹੈ ਅਤੇ y-ਅੰਤਰਖੰਡ ( ੦,੬ ) ( ੦,੬ ) ਹੈ।
x-ਅੰਤਰਖੰਡ ( ੨,੦ ) ( ੨,੦ ) ਹੈ ਅਤੇ y-ਅੰਤਰਖੰਡ ( ੦,−੩ ) ( ੦,−੩ ) ਹੈ।
x-ਅੰਤਰਖੰਡ ( ੩,੦ ) ( ੩,੦ ) ਹੈ ਅਤੇ y-ਅੰਤਰਖੰਡ ( ੦, ੯ ੮ ) ( ੦, ੯ ੮ ) ਹੈ।
y=੪−੨x y=੪−੨x
y= ੫−੨x ੩ y= ੫−੨x ੩
y=੨x− ੪ ੫ y=੨x− ੪ ੫
d= ੭੪ d= ੭੪
36 = 36
62.97 ≈ 62.97
(3, −3/2)
(2, −1)
(0, 0)
y = 0
ਸਮ ਰੇਖੀ ਨਹੀਂ
A: (−3, 2), B: (1, 3), C: (4, 0)
ਕਤਾਰ: x x | y y
ਕਤਾਰ: −3 −3 | 1
ਕਤਾਰ: 0 | 2
ਕਤਾਰ: 3 | 3
ਕਤਾਰ: 6 | 4
ਕਤਾਰ: x | y
ਕਤਾਰ: −3 | 0
ਕਤਾਰ: 0 | 1.5
ਕਤਾਰ: 3 | 3
8.246 = 8.246
5 = 5
(−3, 4)
x = 0, y = −2
x = 0.75, y = 0
x = −1.667, y = 0
15 − 11.2 = 3.8 ਮੀਲ ਛੋਟੀ
6.042 6.042
ਹਰੇਕ ਵਿਕਰਨ ਦਾ ਮੱਧਬਿੰਦੂ ਇੱਕੋ ਬਿੰਦੂ (2,2) ਹੈ। ਨੋਟ ਕਰੋ ਕਿ ਇਹ ਆਇਤਾਂ ਦੀ ਵਿਸ਼ੇਸ਼ਤਾ ਹੈ, ਪਰ ਹੋਰ ਚਤੁਰਭੁਜਾਂ ਦੀ ਨਹੀਂ।
37ਮੀ
54 ਫੁੱਟ
2.2 ਭਾਗ ਅਭਿਆਸ
ਇਸਦਾ ਮਤਲਬ ਹੈ ਕਿ ਉਹਨਾਂ ਦੀ ਢਲਾਣ ਇੱਕੋ ਹੈ।
x ਚੱਲ ਦਾ ਘਾਤ ਅੰਕ 1 ਹੈ। ਇਸਨੂੰ ਪਹਿਲੇ-ਦਰਜੇ ਦਾ ਸਮੀਕਰਨ ਕਿਹਾ ਜਾਂਦਾ ਹੈ।
ਜੇਕਰ ਅਸੀਂ ਕੋਈ ਵੀ ਮੁੱਲ ਸਮੀਕਰਨ ਵਿੱਚ ਪਾਉਂਦੇ ਹਾਂ, ਤਾਂ ਉਹ ਸਮੀਕਰਨ ਵਿੱਚ ਇੱਕ ਅਭਿਵਿਅਕਤੀ ਨੂੰ ਅਨਿਸ਼ਚਿਤ ਬਣਾ ਦਿੰਦੇ ਹਨ (ਡਿਨੋਮੀਨੇਟਰ ਵਿੱਚ ਜ਼ੀਰੋ)।
x=2
x=2/7
x=6
x=3
x=−14
x≠−4; x=−3
x≠1; ਜਦੋਂ ਅਸੀਂ ਇਸਨੂੰ ਹੱਲ ਕਰਦੇ ਹਾਂ ਤਾਂ ਸਾਨੂੰ x=1 ਮਿਲਦਾ ਹੈ, ਜੋ ਕਿ ਬਾਹਰ ਰੱਖਿਆ ਗਿਆ ਹੈ, ਇਸ ਲਈ ਕੋਈ ਹੱਲ ਨਹੀਂ ਹੈ
x≠0; x=−5/2
y=−4/5 x+14/5
y=−3/4 x+2
y=1/2 x+5/2
y=−3x−5
y=7
y=−4
8x+5y=7
ਸਮਾਂਤਰ
Perpendicular
m=− 9 7 m=− 9 7
m= 3 2 m= 3 2
m 1 =− 1 3 , m 2 =3; Perpendicular. m 1 =− 1 3 , m 2 =3; Perpendicular.
y=0.245x−45.662. y=0.245x−45.662. Answers may vary. y min =−50, y max =−40 y min =−50, y max =−40
y=−2.333x+6.667. y=−2.333x+6.667. Answers may vary. y min =−10, y max =10 y min =−10, y max =10
y=− A B x+ C B y=− A B x+ C B
The slope for (−1,1)to (0,4)is 3. The slope for (−1,1)to (2,0)is −1 3 . The slope for (2,0)to (3,3)is 3. The slope for (0,4)to (3,3)is −1 3 . The slope for (−1,1)to (0,4)is 3. The slope for (−1,1)to (2,0)is −1 3 . The slope for (2,0)to (3,3)is 3. The slope for (0,4)to (3,3)is −1 3 .
Yes they are perpendicular.
30 ft
$57.50
220 mi
2.3 Section Exercises
Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.
2,000−x 2,000−x
v+10 v+10
Ann: 23; 23; Beth: 46 46
20+0.05m 20+0.05m
300 min
90+40P 90+40P
6 devices
50,000−x 50,000−x
4 h
She traveled for 2 h at 20 mi/h, or 40 miles.
$5,000 at 8% and $15,000 at 12%
B=100+.05x B=100+.05x
Plan A
R=9 R=9
r= 4 5 r= 4 5 or 0.8
W= P−2L 2 = 58−2(15) 2 =14 W= P−2L 2 = 58−2(15) 2 =14
f= pq p+q = 8(13) 8+13 = 104 21 f= pq p+q = 8(13) 8+13 = 104 21
m= −5 4 m= −5 4
h= 2A b 1 + b 2 h= 2A b 1 + b 2
length = 360 ft; width = 160 ft
405 mi
A=88in . 2 A=88in . 2
28.7
h= V π r 2 h= V π r 2
r= V πh r= V πh
C=12π C=12π
2.4 Section Exercises
Add the real parts together and the imaginary parts together.
Possible answer: i i times i i equals -1, which is not imaginary.
−8+2i −8+2i
14+7i 14+7i
− 23 29 + 15 29 i − 23 29 + 15 29 i
8−i 8−i
−11+4i −11+4i
2−5i 2−5i
6+15i 6+15i
−16+32i −16+32i
−4−7i −4−7i
25
2− 2 3 i 2− 2 3 i
4−6i 4−6i
2 5 + 11 5 i 2 5 + 11 5 i
15i 15i
1+i 3 1+i 3
1 1
−1 −1
128i
( 3 2 + 1 2 i ) 6 =−1 ( 3 2 + 1 2 i ) 6 =−1
3i 3i
0
5−5i 5−5i
−2i −2i
9 2 − 9 2 i 9 2 − 9 2 i
2.5 Section Exercises
It is a second-degree equation (the highest variable exponent is 2).
We want to take advantage of the zero property of multiplication in the fact that if a⋅b=0 a⋅b=0 then it must follow that each factor separately offers a solution to the product being zero: a=0orb=0. a=0orb=0.
One, when no linear term is present (no x term), such as x 2 =16. x 2 =16. Two, when the equation is already in the form (ax+b) 2 =d. (ax+b) 2 =d.
x=6, x=6, x=3 x=3
x= −5 2 , x= −5 2 , x= −1 3 x= −1 3
x=5, x=5, x=−5 x=−5
x= −3 2 , x= −3 2 , x= 3 2 x= 3 2
x=−2,3 x=−2,3
x=0, x=0, x= −3 7 x= −3 7
x=−6, x=−6, x=6 x=6
x=6, x=6, x=−4 x=−4
x=1, x=1, x=−2 x=−2
x=−2, x=−2, x=11 x=11
x=3± 22 x=3± 22
z= 2 3 , z= 2 3 , z=− 1 2 z=− 1 2
x= 3± 17 4 x= 3± 17 4
Not real
One rational
Two real; rational
x= −1± 17 2 x= −1± 17 2
x= 5± 13 6 x= 5± 13 6
x= −1± 17 8 x= −1± 17 8
x≈0.131 x≈0.131 and x≈2.535 x≈2.535
x≈−6.7 x≈−6.7 and x≈1.7 x≈1.7
a x 2 +bx+c = 0 x 2 + b a x = −c a x 2 + b a x+ b 2 4 a 2 = −c a + b 4 a 2 ( x+ b 2a ) 2 = b 2 −4ac 4 a 2 x+ b 2a = ± b 2 −4ac 4 a 2 x = −b± b 2 −4ac 2a a x 2 +bx+c = 0 x 2 + b a x = −c a x 2 + b a x+ b 2 4 a 2 = −c a + b 4 a 2 ( x+ b 2a ) 2 = b 2 −4ac 4 a 2 x+ b 2a = ± b 2 −4ac 4 a 2 x = −b± b 2 −4ac 2a
x(x+10)=119; x(x+10)=119; 7 ft. and 17 ft.
maximum at x=70 x=70
The quadratic equation would be (100x−0.5 x 2 )−(60x+300)=300. (100x−0.5 x 2 )−(60x+300)=300. The two values of x x are 20 and 60.
3 feet
2.6 Section Exercises
This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.
They are probably trying to enter negative 9, but taking the square root of −9 −9 is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in −27. −27.
A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.
x=81 x=81
x=17 x=17
x=8, x=27 x=8, x=27
x=−2,1,−1 x=−2,1,−1
y=0, 3 2 , −3 2 y=0, 3 2 , −3 2
m=1,−1 m=1,−1
x= 2 5 , ±3 i x= 2 5 , ±3 i
x=32 x=32
t= 44 3 t= 44 3
x=3 x=3
x=−2 x=−2
x=4, −4 3 x=4, −4 3
x= −5 4 , 7 4 x= −5 4 , 7 4
x=3,−2 x=3,−2
x=−5 x=−5
x=1,−1,3,-3 x=1,−1,3,-3
x=2,−2 x=2,−2
x=1,5 x=1,5
x ≥ 0 x ≥ 0
x=4,6,−6,−8 x=4,6,−6,−8
10 in.
90 kg
2.7 Section Exercises
When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.
( −∞,∞ ) ( −∞,∞ )
We start by finding the x-intercept, or where the function = 0. Once we have that point, which is (3,0), (3,0), we graph to the right the straight line graph y=x−3, y=x−3, and then when we draw it to the left we plot positive y values, taking the absolute value of them.
( −∞, 3 4 ] ( −∞, 3 4 ]
[ − 13 2 ,∞ ) [ − 13 2 ,∞ )
( −∞,3 ) ( −∞,3 )
( −∞,− 37 3 ] ( −∞,− 37 3 ]
All real numbers ( −∞,∞ ) ( −∞,∞ )
( −∞,− 10 3 )∪( 4,∞ ) ( −∞,− 10 3 )∪( 4,∞ )
( −∞,−4 ]∪[ 8,+∞ ) ( −∞,−4 ]∪[ 8,+∞ )
No solution
( −5,11 ) ( −5,11 )
[ 6,12 ] [ 6,12 ]
[ −10,12 ] [ −10,12 ]
x>−6andx>−2 Take the intersection of two sets. x>−2, (−2,+∞) x>−6andx>−2 Take the intersection of two sets. x>−2, (−2,+∞)
x<−3 or x≥1 Take the union of the two sets. (−∞,−3) ∪ [1,∞) x<−3 or x≥1 Take the union of the two sets. (−∞,−3) ∪ [1,∞)
( −∞,−1 )∪( 3,∞ ) ( −∞,−1 )∪( 3,∞ )
[ −11,−3 ] [ −11,−3 ]
It is never less than zero. No solution.
Where the blue line is above the orange line; point of intersection is x=−3. x=−3.
( −∞,−3 ) ( −∞,−3 )
Where the blue line is above the orange line; always. All real numbers.
(−∞,−∞) (−∞,−∞)
( −1,3 ) ( −1,3 )
( −∞,4 ) ( −∞,4 )
{ x| x<6 } { x| x<6 }
{ x| −3≤x<5 } { x| −3≤x<5 }
( −2,1 ] ( −2,1 ]
( −∞,4 ] ( −∞,4 ]
Where the blue is below the orange; always. All real numbers. (−∞,+∞). (−∞,+∞).
Where the blue is below the orange; ( 1,7 ). ( 1,7 ).
x=2, −4 5 x=2, −4 5
( −7,5 ] ( −7,5 ]
80≤T≤120 1,600≤20T≤2,400 80≤T≤120 1,600≤20T≤2,400
[ 1,600, 2,400 ] [ 1,600, 2,400 ]
Review Exercises
x-intercept: ( 3,0 ); ( 3,0 ); y-intercept: ( 0,−4 ) ( 0,−4 )
y= 5 3 x+4 y= 5 3 x+4
72 =6 2 72 =6 2
620.097 620.097
midpoint is ( 2, 23 2 ) ( 2, 23 2 )
row: x | y
row: 0 | −2
row: 3 | 2
row: 6 | 6
x=4 x=4
x= 12 7 x= 12 7
No solution
y= 1 6 x+ 4 3 y= 1 6 x+ 4 3
y= 2 3 x+6 y= 2 3 x+6
females 17, males 56
84 mi
x=− 3 4 ± i 47 4 x=− 3 4 ± i 47 4
horizontal component −2; −2; vertical component −1 −1
7+11i 7+11i
16i 16i
−16−30i −16−30i
−4−i 10 −4−i 10
x=7−3i x=7−3i
x=−1,−5 x=−1,−5
x=0, 9 7 x=0, 9 7
x=10,−2 x=10,−2
x= −1± 5 4 x= −1± 5 4
x= 2 5 , −1 3 x= 2 5 , −1 3
x=5±2 7 x=5±2 7
x=0,256 x=0,256
x=0,± 2 x=0,± 2
x=−2 x=−2
x= 11 2 , −17 2 x= 11 2 , −17 2
( −∞,4 ) ( −∞,4 )
[ −10 3 ,2 ] [ −10 3 ,2 ]
No solution
( − 4 3 , 1 5 ) ( − 4 3 , 1 5 )
Where the blue is below the orange line; point of intersection is x=3.5. x=3.5.
( 3.5,∞ ) ( 3.5,∞ )
Practice Test
y= 3 2 x+2 y= 3 2 x+2
row: x | y
row: 0 | 2
row: 2 | 5
row: 4 | 8
( 0,−3 ) ( 0,−3 ) ( 4,0 ) ( 4,0 )
( −∞,9 ] ( −∞,9 ]
x=−15 x=−15
x≠−4,2; x≠−4,2; x= −5 2 ,1 x= −5 2 ,1
x= 3± 3 2 x= 3± 3 2
( −4,1 ) ( −4,1 )
y= −5 9 x− 2 9 y= −5 9 x− 2 9
y= 5 2 x−4 y= 5 2 x−4
14i 14i
5 13 − 14 13 i 5 13 − 14 13 i
x=2, −4 3 x=2, −4 3
x= 1 2 ± 2 2 x= 1 2 ± 2 2
4 4
x= 1 2 ,2,−2 x= 1 2 ,2,−2