Michael is making a scale drawings a rectangular rooms using a scale of 1 in ratio 1 and 1/2 feet. He wants to use paper that has a width of 8 and 1/2 inches and a length of 11 inches for the drawings. Determine whether the scale drawings for each of these rims will fit on one piece of paper

Complete question is;

Michael is making scale drawings of rectangular rooms using a scale of 1 inch: 11/5 feet. He wants to use paper that has a width of 8 1/2 inches and a length of 11 inches for the drawings. Determine whether the scale drawings for each of these rooms will fit on one piece of paper.

Choose Yes or No for each set of dimensions.

a. 16 feet by 16 feet Yes/ No

b. 10 feet by 15 feet Yes/ No

c. 15 feet by 20 feet Yes/ No

d. 12 feet by 16 feet Yes/ No

A) Yes

B) Yes

C) Yes

D) Yes

Step-by-step explanation:

We are told that 1 inch: 11/5 feet

This means in decimal form, 1 Inch represents 2.2 feet

A) 16 ft width gives: 16/2.2 inches = 7.27 inches

Also, 16ft length gives; 16/2.2 inches = 7.27 inches

Length and width are both less than the ones given in the question respectively. Thus, it's sufficient.

B) 10 ft width gives; 10/2.2 inches = 4.55 inches

15 ft length gives; 15/2.2 inches = 6.82 inches

Length and width are both less than the ones given in the question respectively. Thus, it's sufficient.

C) 15 ft width gives; 15/2.2 inches = 6.82 inches

20 ft width gives; 20/2.2 inches = 9.09 inches

Length and width are both less than the ones given in the question respectively. Thus, it's sufficient.

D) 12 ft width gives: 12/2.2 inches = 5.45 inches

16ft length gives; 16/2.2 inches = 7.27 inches

Length and width are both less than the ones given in the question respectively. Thus, it's sufficient.

Thus, all the dimensions are sufficient to fit on one piece of paper

e.

step-by-step explanation:

because there are always 10 numbers in a phone number.

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm