Problem 1: find the length of the unknown side of the triangle. a right triangle, one leg is 5 meters, one leg is unknown, and the hypotenuse is 11 meters. the length of the unknown side of the triangle is _ m. (simplify your answer. type an exact answer, using radicals as needed.) tip: a^2+b^2=c^2 problem 2: tip: a^2+b^2=c^2 find the length of the unknown side of each triangle. give the exact length and an approximate length. a right triangle, one leg is 8 millimeters, one leg is 8.1 millimeters, and the hypotenuse is unknown. the exact length of the unknown side is _ mm. (simplify your answer. type an exact answer, using radicals as needed. use integers or decimals for any numbers in the expression.) the approximate length of the unknown side is _ mm. (round to one decimal place as needed.)

Problem #1:

A right triangle, one leg is 5 meters, one leg is unknown, and the hypotenuse is 11 meters.

Use the Pythagorean Theorem, since this is a right triangle. If a and b are sides and c is the hypotenuse, then c^2 = a^2 + b^2.

Subbing the given side lengths, c^2 = (5 m)^2 + (11 m)^2 = 25 m^2 + 121 m^2, or

146 m^2. Thus, the length of the hypotenuse, c, is √(146 m^2), or 12.08 m, to two decimal places.

Problem #2:

A right triangle, one leg is 8 millimeters, one leg is 8.1 millimeters, and the hypotenuse is unknown. Again, use the Pyth. Thm.:

(8.1 mm)^2 + (8 mm)^2 = c^2 = square of the length of the hypotenuse.

Then c^2 = 129.61 mm^2. This is not a perfect square, so we can ony give an approximate value for the hypo length, c: √(129.61 mm^2) or 11.38 mm.

a) $ 9.50

b) 17

c) y-intercept (0, 12). this is the initial amount of money that joe had prior to playing any game.

d) x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e) slope is -0.5. joe's balance will reduce by 0.5 for every game that he plays.

f)

domain; [0,24]

range; [0,12]

g) x + 2y = 24

step-by-step explanation:

a)

if joe plays 5 games, this will imply that x = 5. substitute x = 5 in the equation and solve for y;

y = -.50x + 12

y = -0.50(5) + 12

y = 9.50

b)

if joe leaves the arcade with $3.50, this will mean that y = 3.50. plug in y = 3.50 in the given equation and solve for x;

y = -.50x + 12

3.50 = -0.50x + 12

0.50x = 12 - 3.50

0.5x = 8.5

x = 17

c)

the function given is   y = -.50x + 12.

the y-intercept is the point where the graph of this function will intersect the y-axis. at this point, the value of x is usually 0. plug in x = 0 in the given equation and determine y;

y = -0.50(0) + 12

y = 12

the y-intercept is thus (0, 12). this is the initial amount of money that joe had prior to playing any game.

d)

the x-intercept refers to the point where the graph of a function intersects with the x-axis. at this point, the value of y is 0. substitute y = 0 in the given equation and solve for x;

0 =   -.50x + 12

0.5x = 12

x = 24

x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e)

the slope of the given equation is -0.5. the equation has been given in slope-intercept form. thus the slope is simply the coefficient of x.

joe's balance will reduce by 0.5 for every game that he plays. this is the interpretation of the slope.

f)

domain is the set of x values for which the function is real and defined. the domain will thus be; [0,24]. this will restrict the balance to only assume positive values.

on the other hand, the range is the set of y values for which a function is real and defined. the range in our case will be [0,12]. joe can have between $0 and $12 balance.

g)

the standard form of the equation of a straight line is;

ax + by = c

where a, b, and c are integers.

our equation is given as;

y = -.50x + 12

re-writing the equation we have;

0.5x + y = 12

divide all through by 0.5;

x + 2y = 24

trapezoid

step-by-step explanation:

first i plotted all my points on a grid graph

i made sure all of them were in the correct quadrant

then i saw it was a trapezoid or a quadrilateral