Question

Solve the following multiplication and division problems.
a. 10 yd. × 3 yd.
b. 15 ft. × 18 ft.
c. 13 yd. 2 ft. 6 in. × 15
d. 198 sq. yd. 8 sq. ft. × 7

Answer 1

Answer:

Step-by-step explanation:

a. 10 yd.×3 yd.=30 square yards;

b. 15 ft.×18 ft.=279 square feets;

c. 13 yd.=13×3 ft.=39 ft.=39×12 in.=468 in., 2 ft. 6 in.=2×12+6 in.=30 in., then 13 yd. 2 ft. 6 in.×15=498 in. × 15=7470 in.;

d. 198 sq. yd.=198 ×9 sq. ft.=1782 sq. ft.

198 sq. yd. 8 sq. ft. =1782+8 sq. ft.=1790 sq. ft.

Then 1790 sq. ft. × 7=12530 sq. ft.

Answer 2

Answer:

To solve these problems, we just need to multiply or divide.

To solve problem c and d, we need to transform all units to the same, and then operate.

a.

$10yd \times 3yd = 30yd^{2}$

Remember that $yd \times yd = yd ^{2}$

b.

$15ft \times 18 ft = 270 ft^{2}$

c.

$13yd \ 2ft \ 6in \times 15$

To solve this one, we need to transform all lengths to yards. We know that 1 yard equals 3 feet and 1 feet equals 12 inches. So,

$(13yd + 2ft\frac{1yd}{3ft}+6in\frac{1yd}{36in} \times 15\\(13yd+0.67yd+0.17yd) \times 15 =207.6yd$

d.

$(198yd^{2} + 8ft^{2}) \times 7 \\(198yd^{2} + 8ft^{2}\frac{1yd^{2} }{9ft^{2} }) \times 7 \\(198yd^{2} +0.89yd^{2}) \times 7 =1,392.23 yd^{2}$