The diagram shows a right triangle and three squares. The area of the largest square is 363636 units^2 2 squared. Which could be the areas of the smaller squares?

The answers are A. and B.

Step-by-step explanation:

Since the area of the largest square is 36. We need two numbers that equal 36. and A. had 6 and 30 so i picked it and it was right and B. is 28 and 8 which also equals 36. But, C. is 4 and 16 which is not 36. So A. and B. are the answers. Hope this helps! :)

We can use the Pythagorean theorem (a^2+b^2=c^2)(a  

2

+b  

2

=c  

2

)left parenthesis, a, squared, plus, b, squared, equals, c, squared, right parenthesis to determine possible areas of the two smaller squares.

\text{Area of a square} =\text{side}^2Area of a square=side  

2

start text, A, r, e, a, space, o, f, space, a, space, s, q, u, a, r, e, end text, equals, start text, s, i, d, e, end text, squared

So, we can substitute the areas of the squares that share side lengths with the triangle for a^2, b^2a  

2

,b  

2

a, squared, comma, b, squared and c^2c  

2

c, squared in the Pythagorean theorem.

Hint #22 / 6

For example, in the diagram above, the area of the square that shares a side with the hypotenuse is 363636 square units. So, c^2=36c  

2

=36c, squared, equals, 36.

Hint #33 / 6

Let's fill in the possible values to see if they make the equation true.

\begin{aligned} a^2 + b^2 &= c^2 a^2 + b^2 &= 36 6 + 30 &\stackrel{\large?}{=}36 36 &\stackrel{\checkmark}{=}36 \end{aligned}  

a  

2

+b  

2

a  

2

+b  

2

6+30

36

=c  

2

=36

=

?

36

=

✓

36

The sum of the areas of the squares connected to the two shorter triangle sides is equal to the area of the square connected to the longest side.

So, 666 and 303030 could be the areas of the smaller squares.

Hint #44 / 6

\begin{aligned} a^2 + b^2 &= c^2 a^2 + b^2 &= 36 8 + 28 &\stackrel{\large?}{=}36 36 &\stackrel{\checkmark}{=}36 \end{aligned}  

a  

2

+b  

2

a  

2

+b  

2

8+28

36

=c  

2

=36

=

?

36

=

✓

36

The sum of the areas of the squares connected to the two shorter triangle sides is equal to the area of the square connected to the longest side.

So, 888 and 282828 could be the areas of the smaller squares.

Hint #55 / 6

\begin{aligned} a^2 + b^2 &= c^2 a^2 + b^2 &= 36 4 + 16 &\stackrel{\large?}{=}36 20 &\neq 36 \end{aligned}  

a  

2

+b  

2

a  

2

+b  

2

4+16

20

=c  

2

=36

=

?

36



=36

The sum of the areas of the squares connected to the two shorter triangle sides is not equal to the area of the square connected to the longest side.

So, 444 and 161616 could not be the areas of the smaller squares.

Hint #66 / 6

The area of the smaller squares could be:

666 and 303030

888 and 2828

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)