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Kobotan [32]
3 years ago
11

Nicholas used a \$10$10dollar sign, 10 bills to pay for a granola bar that only costs p dollars.

Mathematics
1 answer:
sladkih [1.3K]3 years ago
4 0

Answer:

\$(10-p)

Step-by-step explanation:

Nicholas used a $10 bill to pay for a granola bar that only costs p dollars.

To find much change Nicholas received, simply subtract the cost of granola bar from the amount of money Nicholas paid:

\$(10-p)

For example, if granola bar costs p=$4, then Nicholas gets

\$(10-4)=\$6

as a change.

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Evaluate 10m+n/4 wnen m=5 and n=16
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Hi!

Let's put the values in the equation.

10 · 5 + 16 ÷ 4 = ?

Using PEMDAS...

Multiplication
50 + 16 ÷ 4 = ?

Division
50 + 4 = ?

Addition
54

The answer is 54

Hope this helps! :)
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Solve the following equation:<br><br> 3 x one half
Kamila [148]
<h3>hello!</h3>

==========================================================

Let's simplify the expression:-

\bigstar{\boxed{\frac{3}{1} *\frac{1}{2}}

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\bigstar{\boxed{\pmb{\frac{3}{2} }}

======================================================

<h3>note:-</h3>

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2 years ago
What is the x intercept of the line 3x+2y-10=0
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5 0
3 years ago
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Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units. What would be the perimeters of the
Dmitriy789 [7]

Answer:

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

Step-by-step explanation:

Given - Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units.

To find -  What would be the perimeters of the other rectangles Orbet could make using only these 24 tiles? How do the areas of the rectangles compare?

Proof -

We know that

Are of Rectangle = Length × Breadth

And Perimeter of Rectangle = 2 (Length + Breadth )

Now,

Given that,

Orbet used 24 square tiles

And perimeter of the rectangle made = 20

So,

Possible Length of rectangle = 6

Possible breadth of Rectangle = 4

Or Vice-versa.

Total number of Rectangles possible = 4

Possibilities are -  1 × 24, 2 × 12, 3 × 8, 4 × 6

Case I :

Length of rectangle = 1

Breadth of rectangle = 24

∴ Perimeter of rectangle = 2(1 + 24) = 2(25) = 50 units

Area of rectangle = 1 × 24 = 24 unit²

Case II :

Length of rectangle = 2

Breadth of rectangle = 12

∴ Perimeter of rectangle = 2(2 + 12) = 2(14) = 28 units

Area of rectangle = 2 × 12 = 24 unit²

Case III :

Length of rectangle = 3

Breadth of rectangle = 8

∴ Perimeter of rectangle = 2(3 + 8) = 2(11) = 22 units

Area of rectangle = 3 × 8 = 24 unit²

Case IV :

Length of rectangle = 4

Breadth of rectangle = 6

∴ Perimeter of rectangle = 2(4 + 6) = 2(10) = 20 units

Area of rectangle = 4 × 6 = 24 unit²

∴ we get

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

5 0
3 years ago
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