find the area and perimeter of a rectangle with a width of 2 1/3inch and a length of 5 2/5inches. show your work please.

2 answers:

5 0

A=LxW, but first you should convert the mixed fractions to improper fractions:
7/3 x 27/5 = 189/15 = 12 6/10 = 12 3/5 square inches

P = L+L+W+W here I would convert to improper fractions, but then you need to find a common denominator.
P = 7/3 + 7/3 + 27/5 + 27/5 = 14/3 + 54/5 = 70/15 + 162/15 = 232/15 = 15 7/15 inches

aleksandrvk [35]2 years ago

4 0

Area: If you multiply 2 1/3 by 5 2/5, you get an answer of about: 12.6, which means the area is about: 12.6in. squared.
Perimeter: If you add 2 1/3+ 2 1/3+ 5 2/5+ 5 2/5 you get an answer of about: 15.47. So, the perimeter is about 15.47in.

Note: If you aren't allowed to use a calculator on this test, you should probably multiply the numbers yourself so you get show the work and get the answer in a fraction, which is probably what they want. If you are allowed to use calculators, all the information you need is right here.

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substract 6 from me. then multiply by 2. if you substract 40 and then divide by 4. you get 8. what number am i?

stich3 [128]

8x4=32
32+40=72
72/2=36
36+6=42

The number is 42

6 0

3 years ago

Help, please?? ^ω^ xx

Rudiy27

Step 1. Use Multiplication Distributive Property
4*3^3(x^2)^3(y^4)^3
Step 2. Simplify 3^3 to 27
4*27(x^2)^3(y^4)^3
Step 3. Use the Power Rule
4*27x^6(y^4)^3
Step 4. Use the Power Rule
4*27x^6y^12
Step 5. Simplify.
The answer is 108x^6y^12

7 0

3 years ago

Consider the expansion of (5p + 2q)^6. Determine the coefficients for the terms with the powers of p and q shown.

Step2247 [10]

Answer:

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