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inn [45]
2 years ago
4

Solve the equation 5x + (−2) = 6x + 4 using the algebra tiles. What tiles need to be added to both sides to remove the smaller x

-coefficient? What tiles need to be added to both sides to remove the constant from the right side of the equation? What is the solution
Mathematics
2 answers:
Tanzania [10]2 years ago
6 0

Answer:-6

Step-by-step explanation:

5x+(-2)=6x+4

5x-2=6x+4

Add 2 to both sides

5x-2+2=6x+4+2

5x=6+6x

subtract 5x from both sides

5x-5x=6+6x-5x

0=6+x

Subtract 6 from both sides

0-6=6-6+x

-6=x

x=-6

asambeis [7]2 years ago
5 0

Answer:

What tiles need to be added to both sides to remove the smaller x-coefficient?

✔ 5 negative x-tiles

What tiles need to be added to both sides to remove the constant from the right side of the equation?

✔ 4 negative unit tiles

What is the solution?

✔ x = –6

Step-by-step explanation:

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Hi there! Osvoldo did not meet his goal.

(In my answering I suppose you mean 30%, 220 grams of carbohydrates and 55 grams of whole grains, if this is incorrect, please let me know)

Today Osvoldo ate 220 grams of carbohydrates.
10 % of 220 is 22 (divide by 10), and therefore 30 % of 220 is 22 * 3 = 66. 

If at least 66 grams of Osvoldo's total consumption consisted of whole grains, he would have met his goal. However, he only ate 55 grams of whole grains (which is less than 66), and therefore he did not meet his goal.
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2 years ago
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Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val
Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

A(t) = Pe^{rt}

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

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2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

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2009 is 10 years after 1999, so this is A(10).

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The value of the account in the year 2009 will be $682.

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Answer:

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The method of reduction of order is applicable for second-order differential equations.

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Check attachments for the solution to this problem.

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