Answer:
A) Independent variable is "c"
Dependent variable is "p"
B) Yes, it's a function
C) p = 7c - 31
D)domain of the function: 0 ≤ c ≤ 40
Range of the function: 0 ≤ p ≤ 249
Step-by-step explanation:
A) c is the total number of cars washed and p is the profit. The profit depends on the total number of cars washed. Hence, profit(p) is the dependent variable while "c" is independent variable
B) Yes it's a function because for each number of cars washed, there is a distinct value of profit made.
C) We are told that the liquid soap costs $31 and is enough to wash 40 cars. Each car is charged 7$.
$31 is cost to wash one car. Thus profit made per car is;
p = 7c - 31
D) The cost of $31 liquid soap is enough to only wash a maximum of 40 cars. This means the domain for the cars is from 0 cars to 40 cars
Thus;
domain of the function is 0 ≤ c ≤ 40
The profit function is given in answer C above as p = 7c - 31.
Now, if 0 cars are washed they will make zero profit but if 40 cars are washed then;
p = 7(40) - 31
p = 280 - 31
p = 249
Range of the function: 0 ≤ p ≤ 249
The equation y= 2
has one real root and that is x=-1.
What is real roots of the equation?
We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation 
-7x+12=0 is solved, the actual roots are 3 and 4.
Here given,
=> y = 2
Take y=0 then,
=> 2=0
=> =0
=>(x+1)=0
=> x=-1
Hence the given equation has one real root and that is x=-1.
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Answer:
I think it's B
Step-by-step explanation:
Let <em>a</em> and <em>b</em> be the zeroes of <em>x</em>² + <em>kx</em> + 12 such that |<em>a</em> - <em>b</em>| = 1.
By the factor theorem, we can write the quadratic in terms of its zeroes as
<em>x</em>² + <em>kx</em> + 12 = (<em>x</em> - <em>a</em>) (<em>x</em> - <em>b</em>)
Expand the right side and equate the coefficients:
<em>x</em>² + <em>kx</em> + 12 = <em>x</em>² - (<em>a</em> + <em>b</em>) <em>x</em> + <em>ab</em>
Then
<em>a</em> + <em>b</em> = -<em>k</em>
<em>ab</em> = 12
The condition that |<em>a</em> - <em>b</em>| = 1 has two cases, so without loss of generality assume <em>a</em> > <em>b</em>, so that |<em>a</em> - <em>b</em>| = <em>a</em> - <em>b</em>.
Then if <em>a</em> - <em>b</em> = 1, we get <em>b</em> = <em>a</em> - 1. Substitute this into the equations above and solve for <em>k</em> :
<em>a</em> + (<em>a</em> - 1) = -<em>k</em> → 2<em>a</em> = 1 - <em>k</em> → <em>a</em> = (1 - <em>k</em>)/2
<em>a</em> (<em>a</em> - 1) = 12 → (1 - <em>k</em>)/2 • ((1 - <em>k</em>)/2 - 1) = 12
→ (1 - <em>k</em>)²/4 - (1 - <em>k</em>)/2 = 12
→ (1 - <em>k</em>)² - 2 (1 - <em>k</em>) = 48
→ (1 - 2<em>k</em> + <em>k</em>²) - 2 (1 - <em>k</em>) = 48
→ <em>k</em>² - 1 = 48
→ <em>k</em>² = 49
→ <em>k</em> = ± √(49) = ±7
The summation notation for the series 500+490+480+ . . . . . . . +20+10 is
A summation notation is used to express a long summation into a single notation.
The given series is:
500+490+480+ . . . . . . . +20+10
This is an arithmetic series with:
a(1) = 500, and
d = 490 - 500 = -10
The last term is a(n) = 10. Find n first by using the nth term formula of an arithmetic sequence:
a(n) = a(1) + (n-1) . d
10 = 500 + (n-1) . (-10)
10 = 500 -10n + 10
10 n = 500
n = 50
Write the explicit formula for the nth term:
a(n) = a(1) + (n-1) . d
a(n) = 500 + (n-1) . (-10)
a(n) = 500 -10n + 10
a(n) = 510 - 10n
The series is the summation notation from n=1 to n = 50
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