Question

HIII DOES ANYBODY KNOW HOW TO DO #’s 9, 10, and 11? THANK U SM!!!!❤️❤️❤️

Answer 1

Answer:

9. $A = 4608~in.^2$

10. $V = \sqrt{5}~ft^3$

11. (a) $2\sqrt{6}~ft$

11. (b) $perimeter = (6 + 2\sqrt{6})~ft$

11. (c) $area = 3~ft^2$

Step-by-step explanation:

9. Area of square:

$A = s^2$

$A = (48\sqrt{2}~in.)^2$

$A = 48^2 \times (\sqrt{2})^2~in.^2$

$A = 2304 \times 2~in.^2$

$A = 4608~in.^2$

10. Volume of prism:

V = LWH

$V = LWH$

$V = \sqrt{5} \times (2 + \sqrt{3}) \times (2 - \sqrt{3})~ft^3$

The last two factors are a sum and a difference, so I will multiply them first. The product of a sum and a difference is the difference of two squares.

$V = \sqrt{5} \times (4 - 3)~ft^3$

$V = \sqrt{5} \times 1~ft^3$

$V = \sqrt{5}~ft^3$

11.

(a) Use the Pythagorean theorem.

$a^2 + b^2 = c^2$

$c^2 = a^2 + b^2$

$c^2 = (3 + \sqrt{3})^2 + (3 - \sqrt{3})^2$

$c^2 = 9 + 6\sqrt{3} + 3 + 9 - 6\sqrt{3} + 3$

$c^2 = 9 + 3 + 9 + 3 + 6\sqrt{3} - 6\sqrt{3}$

$c^2 = 24$

$c = \sqrt{24}$

$c = \sqrt{4 \times 6}$

$c = 2\sqrt{6}~ft$

(b) perimeter = sum of lengths of three sides

$perimeter = 3 + \sqrt{3} + 3 - \sqrt{3} + 2\sqrt{6}$

$perimeter = (6 + 2\sqrt{6})~ft$

(c) area = base * height/2

The base and height are the legs.

$area = \dfrac{bh}{2}$

$area = \dfrac{(3 + \sqrt{3})(3 - \sqrt{3})}{2}$

$area = \dfrac{9 - 3}{2}$

$area = \dfrac{6}{2}$

$area = 3~ft^2$