<span><span>Solve <span>x5 + 3x4 – 23x3 – 51x2 + 94x + 120 </span></span>><span><span> 0</span>. </span></span><span>First, I factor to find the zeroes:<span><span>x5 + 3x4 – 23x3 – 51x2 + 94x + 120</span><span>= (x + 5)(x + 3)(x + 1)(x – 2)(x – 4) = 0</span></span><span>...so </span><span>x = –5, –3, –1, 2,</span><span> and </span>4<span> are the zeroes of this polynomial. (Review how to </span>solve polynomials, if you're not sure how to get this solution.)<span>To solve by the Test-Point Method, I would pick a sample point in each interval, the intervals being </span>(negative infinity, –5)<span>, </span>(–5, –3)<span>, </span>(–3, –1)<span>, </span>(–1, 2)<span>, </span>(2, 4)<span>, and </span>(4, positive infinity). As you can see, if your polynomial or rational function has many factors, the Test-Point Method can become quite time-consuming.<span>To solve by the Factor Method, I would solve each factor for its positivity: </span><span>x + 5 > 0</span><span> for </span><span>x > –5</span>;<span>x + 3 > 0</span><span> for </span><span>x > –3</span><span>; </span><span>x + 1 > 0</span><span> for </span><span>x > –1</span><span>; </span><span>x – 2 > 0</span><span> for </span><span>x > 2</span><span>; and </span><span>x – 4 > 0</span><span> for </span><span>x > 4</span>. Then I draw the grid:...and fill it in:...and solve:<span>Then the solution (remembering to include the endpoints, because this is an "or equal to" inequality) is the set of </span>x-values in the intervals<span> [–5, –3]<span>, </span>[–1, 2]<span>, and </span>[4, positive infinity]</span>. </span>
As you can see, if your polynomial or rational function has many factors, the Factor Method can be much faster.
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Answer:
M = 6.50n
Step-by-step explanation:
since it is per child, it will be multiplied by the amount of children, n. Therefore giving us the cost of M, or the equivalent.
Answer:
12/5 or 2 2/5
Step-by-step explanation:
3/5+3/5+3/5+3/5
=> (3+3+3+3) / 5
=> 12/5
=> 2 2/5
Hope it helps......
Answer:
6
Step-by-step explanation:
Suppose the width is "a". Then the length is 2a.
2(a²) =72 Divide both sides by 2
a²= 36
a=6
Problem 3: Let x = price of bag of pretzels Let y = price of box of granola bars
We have Lesley's purchase: 4x+2y=13.50
And Landon's: 1x+5y=17.55
We can use the elimination method. Let's negate Landon's purchase by multiplying by -1. -1x-5y=-17.55
We add this four times to Lesley's purchase to eliminate the x variable.
2y-20y=13.50-70.2
-18y=-56.7
y = $3.15 = Price of box of granola bars
Plug back into Landon's purchase to solve for pretzels.
x+5*3.15=17.55
x+15.75=17.55
x = $1.80 = price of bag of pretzels
Problem 4.
Let w = number of wood bats sold
Let m = number of metal bats sold
From sales information we have: w + m = 23
24w+30m=606
Substitution works well here. Solve for w in the first equation, w = 23 - m, and plug this into the second.
24*(23-m)+30m=606
552-24m+30m=606
6m=54
m=9 = number of metal bats sold
Therefore since w = 23-m, w = 23-9 = 14. 14 wooden bats were sold.
