Answer:
1/2 + 5/2 i
Step-by-step explanation:
(-5+i) i
---------* --------
2i i
cannot have i in the denominator so we multiply by i
(-5+i) i -5i + i^2
---------* -------- = ----------------
2i i +2i^2
remember i^2 = -1
-5i + -1 -1 -5i
= ---------------- = ------------------
+2(-1) -2
get rid of the negative in the denominator by multiplying the top and bottom by -1
(-1 -5i) * -1
= ------------------
-2*(-1)
(1 +5i)
= ------------------ = 1/2 + 5/2 i
2
$100x should be greater than or equal to $2500. X being the number of computers sold. If you divide 100 by 2500 you will know how many computers he need to sell.
$100x >(or equal to) $2500
<u>vertex</u>
y = 3(x - 2)² - 4
y = 3((x - 2)(x - 2)) - 4
y = 3(x² - 2x - 2x + 4) - 4
y = 3(x² - 4x + 4) - 4
y = 3(x²) - 3(4x) + 3(4) - 4
y = 3x² - 12x + 12 - 4
y = 3x² - 12x + 8
3x² - 12x + 8 = 0
x = <u>-(-12) +/- √((-12)² - 4(3)(8))</u>
2(3)
x = <u>12 +/- √(144 - 96)</u>
6
x = <u>12 +/- √(48)
</u> <u> </u> 6<u>
</u>x =<u> 12 +/- 6.93</u>
<u /> 6
x = 2 +/- 1.155
x = 2 + 1.155 x = 2 - 1.155
x = 3.155 x = 0.845
y = 3x² - 12x + 8
y = 3(3.155)² - 12(3.155) + 8
y = 3(1.334025) - 3.786 + 8
y = 4.002075 - 3.786 + 8
y = 0.216075 + 8
y = 8.216075
(x, y) = (3.155, 8.216075)
or
y = 3x² - 12x + 8
y = 3(0.845)² - 12(0.845) + 8
y = 3(0.714025) - 10.14 + 8
y = 2.142075 - 10.14 + 8
y = -7.857925 + 8
y = 0.142675
(x, y) = (0.845, 0.142675)
<u>y-intercept</u>
y = 3x² - 12x + 8
y = 3(0)² - 12(0) + 8
y = 3(0) - 0 + 8
y = 0 - 0 + 8
y = 0 + 8
0 = -y + 8
y = 8
(x, y) = (0, 8)
-------------------------------------------------------------------------------------------
<u>vertex</u>
y = 4(x - 5)² = 1
y = 4(x - 5)² - 1
y = 4((x - 5)(x - 5)) - 1
y = 4(x² - 5x - 5x + 25) - 1
y = 4(x² - 10x + 25) - 1
y = 4(x²) - 4(10x) + 4(25) - 1
y = 4x² - 40x + 100 - 1
y = 4x² - 40x + 99
4x² - 40x + 99 = 0
x = <u>-(-40) +/- √((-40)² - 4(4)(99))</u>
2(4)
x = <u>40 +/- √(1600 - 1584)</u>
8
x = <u>40 +/- √(16)</u>
8
x = <u>40 +/- 4</u>
8
x = 5 +/- 1/2
x = 5 + 1/2 x = 5 - 1/2
x = 5 1/2 x = 4 1/2
y = 4x² - 40x + 99
y = 4(5 1/2)² - 40(5 1/2) + 99
y = 4(30 1/4) - 220 + 99
y = 121 - 220 + 99
y = -99 + 99
y = 0
(x, y) = (5 1/2, 0)
or
y = 4x² - 40x + 99
y = 4(4 1/2)² - 40(4 1/2) + 99
y = 4(20 1/4) - 180 + 99
y = 81 - 180 + 99
y = -99 + 99
y = 0
(x, y) = (4 1/2, 0)
<u>y-intercept</u>
y = 4x² - 40x + 99
y = 4(0)² - 40(0) + 99
y = 4(0) - 0 + 99
y = 0 - 0 + 99
y = 0 + 99
y = 99
(x, y) = (0, 99)
--------------------------------------------------------------------------------------------
<u>vertex</u>
y = (x - 1)² = 2
y = (x - 1)² - 2
y = ((x - 1)(x - 1)) - 2
y = (x² - x - x + 1) - 2
y = x² - 2x + 1 - 2
y = x² - 2x - 1
x² - 2x - 1 = 0
x = <u>-(-2) +/- √((-2)² - 4(1)(-1))</u>
2(1)
x = <u>2 +/- √(4 + 4)</u>
2
x = <u>2 +/- √(8)</u>
2
x = <u>2 +/- 2.83</u>
2
x = 1 +/- 1.415
x = 1 + 1.415 x = 1 - 1.415
x = 2.415 x = 0.415
y = x² - 2x - 1
y = (2.145)² - 2(2.145) - 1
y = 4.60125 - 4.029 - 1
y = 0.57225 - 1
y = 0.42775
(x, y) = (2.415, 0.42775)
or
y = x² - 2x - 1
y = (0.415)² - 2(0.415) - 1
y = 0.172225 - 0.83 - 1
y = -0.657775 - 1
y = -1.657775
(x, y) = (0.415, -1.657775)
<u>y-intercept</u>
y = x² - 2x - 1
y = (0)² - 2(0) - 1
y = 0 - 0 - 1
y = 0 - 1
y = -1
(x, y) = (0, -1)
<h2>Solving Equations</h2>
To solve linear equations, we must perform inverse operations on both sides of the equal sign to <em>cancel values out</em>.
- If something is being added to x, subtract it from both sides.
- If something is being subtracted from x, add it on both sides.
- Same with multiplication and division. If x is being divided, multiply. If x is being multiplied, divide.
We perform inverse operations to<em> combine like terms</em>. This means to get x to one side and everything else on the other.
<h2>Solving the Questions</h2><h3>Question 1</h3>
Because 7 is being added to x, subtract it from both sides:
Because x is being multiplied by 5, divide both sides by 5:
Therefore. 
.
<h3>Question 2</h3>
Here, we can group all the x values on the left side of the equation. Subtract 5x from both sides:
To isolate x, subtract 4 from both sides:
Divide both sides by 2:
Therefore, 
.
