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seropon [69]
3 years ago
13

How can you use unit rates to solve pricing problems?​

Mathematics
2 answers:
melomori [17]3 years ago
5 0

Answer:

a ratio is a comparison of two numbers or measurements

the objects being compared are called the terms of the ratio.

if the tow terms are different units this is called a rate.

example:

David covers 600 miles in 8 hours and John covers 380 miles in 5 hours. Who is driving faster ?

solution:

Distance covered by David in 1 hour is

=  600 / 8  

=  75 miles per hour

Distance covered by John in 1 hour is

=  380 / 5  

=  76 miles per hour

When we compare the above unit rates (Distance covered in 1 hour), John is driving faster than David.

Llana [10]3 years ago
3 0

We can use unit rates to solve pricing problems because we solve for price of one of that item.

For example, David bought 5 pounds of peanuts for $10, how much does it cost for one pound?

10 / 5 = $2 per pound

The $2 is the unit rate and we use to find if David buys more then 10 pounds.

Best of Luck!

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What is the absolute value of <br> |-20|+|30|
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3 years ago
What is the distance between the ordered pairs (-5.5, -3) and (6.5, -3)
motikmotik

Step-by-step explanation:

√(-5.5-6.5)²+(-3+3)²

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4 0
3 years ago
Read 2 more answers
HELP, I AM UTTERLY CONFUSED *20 points
Studentka2010 [4]
Ok so this question is a bit complicated, but it's easier to understand if you break it down into smaller parts!

1) First, you know that ABGF is half the perimeter of ACDE. This means that the length of one side of ABGF must be 1/2 the length of one side of ACDE.
>> You can think of this by putting in random numbers. Say the perimeter of the larger square is 24 and the perimeter of the smaller square is 12. That means one side of the larger square of 24/4 (b/c four sides) = 6 and one side of the smaller square is 12/4 = 3!

2. Ok know you know the lengths of the sides relative to each other, but you're only given one value: 4in. Since the smaller square has sides that are 1/2 the larger squares, you know that it makes up 1/4 of the larger square! So imagine 4 of those smaller squares filling up that larger square to make a 2 by 2. It just so happens that 4in is the diagonal going through one of our imaginary squares, which is equal in size to ABGF!

3. Now use the 45-45-90 rule to figure out the length of one side of that imaginary square because the 4in diagonal splits that imaginary square into two of those 45-45-90 triangles. You know the hypotenuse of that triangle is 4in. That means one of the legs is 4/✓2 (since the rule says that the hypotenuse and the leg are in a ✓2:1 ratio). And like we said before the length of that leg is the length of the side of our imaginary square. And our imaginary square must be the same size as ABGF! So now we know the side of the smaller square to be 4/✓2!

4. Multiply the side of the smaller square by 2 to get the side of our larger square. (4/✓2)*2=8/✓2

5. Now to find the area of the shaded region, just find the area of the smaller square ABGF and subtract from the larger square ACDE. Use equation for the area of a square!
a =  {s}^{2}
where s=the length of one side.

The length of one side of the smaller square is 4/✓2. So it's area is:
{( \frac{4}{ \sqrt{2} }) }^{2}  =  \frac{16}{2}  = 8

The length of one side of the larger square is 8/✓2. So it's area is:
{ ( \frac{8}{ \sqrt{2} }) }^{2}  =   \frac{64}{2}   = 32

Now subtract. 32-8=24! :)

Hope this helps! Let me know if you have any questions.
3 0
4 years ago
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