"a student made the following chart to prove that ab2 + bc2 = ac2.
statement justification
1. triangle abc is similar to triangle bdc 1. angle abc = angle bdc and angle bca = angle bcd
2. bc2 = ac × dc 2. bc ÷ dc = ac ÷ bc because triangle abc is similar to triangle bdc
3. triangle abc is similar to triangle abd 3. angle abc = angle bad and angle bac = angle abd
4. ab2 = ac × ad 4. ab ÷ ad = ac ÷ ab because triangle abc is similar to triangle abd
5. ab2 + bc2 = ac × ad + ac × dc
= ac (ad + dc)
6. adding statement 1 and statement 2
7. ab2 + bc2 = ac2
which of the first four steps is the error? "

Solution:

Given : In triangle ABC right angled at B.

To Prove :

Construction: Draw B D perpendicular AC.

Proof: 1. Triangle ABC is similar to triangle BDC 1. Angle ABC = Angle BDC and Angle BCA = Angle BCD  

2. = AC × DC 2. BC ÷ DC = AC ÷ BC because triangle ABC is similar to triangle BDC  

3. Triangle ABC is similar to triangle ABD 3. Angle ABC = Angle BAD and Angle BAC = Angle ABD  

4.  = AC × AD 4. AB ÷ AD = AC ÷ AB because triangle ABC is similar to triangle ABD  

5.

= AC (AD + DC)  

6. Adding Statement 1 and Statement 2  

7.

→→→Statement 3 is incorrect.It should be replaced by, Angle ABC = Angle ADB and Angle B AC = Angle BAD .

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

step-by-step explanation:

1