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Sliva [168]
3 years ago
12

By what can you divide 66 and 72

Mathematics
1 answer:
pashok25 [27]3 years ago
7 0

Answer:

You can divide them both by two and three.


Step-by-step explanation:


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Figure #5 shows 5 squares which I would assume means (5, 5)
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x     y
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3 years ago
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5. fourth or \frac{5x-15}{x^{2}-3x-28}
6. third or\frac{4x+25}{x^{2}+5x+4}
7.the answe ris x+3 since it can be factored out of the 5x+15 tingie (second )
8. fourth or \frac{x^{2}+4x}{4}
9. fourth or \frac{2y+6}{9y-7} times \frac{3}{2}
10. third or \frac{(x+3)(x+8)}{(x-8)(x-8)}
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5 0
2 years ago
The two figures shown are congruent. Which statement is true?
zimovet [89]

9514 1404 393

Answer:

  (b) One figure is a rotation image of the other

Step-by-step explanation:

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6 0
3 years ago
where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 ≤ x ≤
marusya05 [52]

THIS IS THE COMPLETE QUESTION BELOW

The demand equation for a product is p=90000/400+3x where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 ≤ x ≤ 50.

Answer

$168.27

Step by step Explanation

Given p=90000/400+3x

With the limits of 40 to 50

Then we need the integral in the form below to find the average price

1/(g-d)∫ⁿₐf(x)dx

Where n= 40 and a= 50, then if we substitute p and the limits then we integrate

1/(50-40)∫⁵⁰₄₀(90000/400+3x)

1/10∫⁵⁰₄₀(90000/400+3x)

If we perform some factorization we have

90000/(10)(3)∫3dx/(400+3x)

3000[ln400+3x]₄₀⁵⁰

Then let substitute the upper and lower limits we have

3000[ln400+3(50)]-ln[400+3(40]

30000[ln550-ln520]

3000[6.3099×6.254]

3000[0.056]

=168.27

the average price p on the interval 40 ≤ x ≤ 50 is

=$168.27

8 0
3 years ago
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