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Stells [14]
4 years ago
12

Derrick's Dog Sitting and Darlene's Dog Sitting are competing for new business. The companies ran the ads shown. A. Write and so

lve an equation to find the number of hours for which the total cost will be the same for the two services
Mathematics
1 answer:
LuckyWell [14K]4 years ago
7 0
All you have to do is for example d+k
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Using the 4 train car strategy for factoring polynomials, what is the green common factor for the 1st two train cars and the gre
koban [17]

The completely factored expression of 2x^2 + 4x + 3xy + 6y is (2x + 3y)(x + 2)

<h3>How to factor the polynomial?</h3>

The expression is given as:

2x^2 + 4x + 3xy + 6y

Group the expression into two

[2x^2 + 4x] + [3xy + 6y]

Factor out each group

2x(x + 2) + 3y(x + 2)

Factor out x + 2

(2x + 3y)(x + 2)

Hence, the completely factored expression of 2x^2 + 4x + 3xy + 6y is (2x + 3y)(x + 2)

Read more about factored expression at:

brainly.com/question/723406

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7 0
2 years ago
Drag and drop the answer into the box to identify each function as linear, exponential, or neither.
leonid [27]

Yes these are correct hope it helps.    


8 0
4 years ago
Read 2 more answers
Factorize x^3-2x^2-13x-10
Alecsey [184]
X^3-2x^2-13x-10=(x+1)(x+2)(x-5)
4 0
3 years ago
In kickboxing, it is found that the force, f, needed to break a board, varies inversely with the length, l, of the board. If it
GalinKa [24]

Answer:

4.2

Step-by-step explanation:

f varies inversly with L can be translated matimatically as:

● f = k/L

It takes 7 pounds of pressure to break a 3 feet long board.

Replace f by 7 and L by 3.

● 7 = k/3 => k=7×3=21

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's find tge length of a board that takes 5 pounds of pressure to be broken.

● 5 = k/L

● 5 = 21/L

● L = 21/5 = 4.2

So the board is 4.2 feet long

3 0
3 years ago
The value of log3 5 x log25 9 is
Rama09 [41]
\log_{3}5\cdot\log_{25}9=\dfrac{1}{\log_{5}3}\cdot\log_{25}9=\dfrac{1}{\log_{5}3}\cdot\log_{5^2}9=\dfrac{1}{\log_{5}3}\cdot\dfrac{1}{2}\log_{5}9=\\\\\\=&#10;\dfrac{1}{\log_{5}3}\cdot\dfrac{1}{2}\log_{5}3^2=\dfrac{1}{\log_{5}3}\cdot\dfrac{2}{2}\log_{5}3=\dfrac{1}{\log_{5}3}\cdot\log_{5}3=\dfrac{\log_{5}3}{\log_{5}3}=\boxed{1}
8 0
3 years ago
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