Explore forming triangles with different combinations of side lengths.

✓ 1. Try to form a triangle by dragging

points A and D. Is it possible to form a

triangle with side lengths 1, 5, and 3

units? no

2. Is it possible to form a triangle with

side lengths 3, 4, and 5 units?

AB = 3 units

BC = 4 units

CD = 5 units

UUT

(A) No

(B) Yes

Step-by-step explanation:

(A) Using centimeters for example, it is not possible to create a closed or full triangle with side lengths 1cm, 5cm and 3cm. [See attachment for illustration]

(B) It is possible to create a triangle of side lengths 3cm, 4cm and 5cm. [See attachment also]

The Pythagoras Rule also applies in this case.

The sum of the squares of the lengths of the 2 shorter sides (the opposite side and the adjacent side) is equal to the square of the longest side (the hypotenuse).

(3×3) + (4×4) = (5×5)

9 + 16 = 25

1) No

2) Yes

3) No

4) Yes

5) Yes

Step-by-step explanation:

edge 2020:)

All correct

d. vertex

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?