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Papessa [141]
3 years ago
13

Jordan is a manager of a car dealership. He has two professional car washers, Matthew and Arianna, to clean the entire lot of ca

rs. Matthew can wash all the cars in 14 hours. Arianna can wash all the cars in 11 hours. Jordan wants to know how long it will take them to wash all the cars in the lot if they work together.
Mathematics
1 answer:
Maslowich3 years ago
8 0
If we say there are 100 cars in a lot (It's an arbitrary number. You can use any number you want), Matthew can wash 100 cars in 14 hours or 7.14 cars an hour. Arianna can wash 100 cars in 11 hours or 9.09 cars an hour. Together, in an hour they can wash 7.14 + 9.09 or 16.23 cars. So to wash 100 cars it would take them 100/16.23 = 6.16 or about 6 hours. 
You can do this with variables as well. 
Matthew has a rate of x/14 while Arianna has a rate of x/11. Together their rate is x/14 + x/11= x/y
Divide both sides by x, you would get 1/14 + 1/11 = 1/y
Solve for y which would be around 6 hours. 
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Answer:

<h2>12 4/8</h2>

Step-by-step explanation:

1. First step

ADD

7+4=11

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so

11\frac{12}{8}

2. Second step

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Evaluate : -80 - (-15) =
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Find the 2th term of the expansion of (a-b)^4.​
vladimir1956 [14]

The second term of the expansion is -4a^3b.

Solution:

Given expression:

(a-b)^4

To find the second term of the expansion.

(a-b)^4

Using Binomial theorem,

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here, a = a and b = –b

$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}

Substitute i = 0, we get

$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4

Substitute i = 1, we get

$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b

Substitute i = 2, we get

$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}

Substitute i = 3, we get

$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}

Substitute i = 4, we get

$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}

Therefore,

$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}

=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}

Hence the second term of the expansion is -4a^3b.

3 0
3 years ago
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