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amm1812
3 years ago
10

Antonio buys a 10-pound bag of potatoes for $3.98. To the nearest cent, how much is 1 pound of potatoes?

Mathematics
1 answer:
valentinak56 [21]3 years ago
3 0

Answer:

C $0.40

Step-by-step explanation:

to find how much one pound of potatoes is, you have to divide the the cost of the bad by the number of pounds ($3.98 divided by 10 pounds)

$3.98 : 10 = $0.398 = $0.40

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Answer:

1:5

Step-by-step explanation:

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Select all that apply.
Mila [183]

First we need to have same units before comparing them. That is, either both of them are in cm or both of them are in mm. So if we need to convert 12.5 cm to mm, we know that 1 cm =10 mm. So we have to find out how many mm are in 12.5 cm. And let 12.5 cm =x mm. So to find the value of x , we set a proportion and solve for x. That is

\frac{1cm}{10mm}=\frac{12.5cm}{x mm}

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3 years ago
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3 years ago
What are the zeros of the polynomial function? <br><br> F(x)= x^2 +12x
Pavel [41]

Answer:

x = 0 and x = -12 are the two zeroes

Step-by-step explanation:

"Zeroes" of a function are also called the solutions, roots or x intercepts.

The zeroes are the numbers that give the function a values of 0, so find what value for 'x' yields F(x) = 0 as a result.

With quadratic equations (equations with x² as the largest exponent), you factor the equation.  

Step 1: Set the equation equal to zero

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 Depending on the equation, there are several methods you can use to factor, for this one we can factor out an 'x' since both terms have an 'x' in them.  We get...

  x(x + 12) = 0      

So now, using the "zero product property" which means that if two things are multiplied together, and the result is zero, then either the first term is zero, the second term is zero, or they are both zero.  So we set each term equal to zero and solve.

So

either    x = 0     or        x + 12 = 0

Solve each for x to get your two anwers.

x= 0 is already solved, so 0 is our first zero

Now solve

x + 12 = 0  

  x = -12             (subtract 12 from both sides to isolate x)

 

8 0
3 years ago
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