Desmos Introduction to Geometry Proofs Proof #5 Challenge
1 answer:
Answer:
For statement 2, reason is Segment addition postulate.
Statement 3 and reason is
BD=AC+CD using substitution property of equality.
Statement 4 and reason is
AC+CD=BD using symmetric property of equality
Statement 5 is
BD= BC+CD
Statement 6 and reason is
AC+CD=BC+CD using transitive property of equality
Reason for 7 is
Subtraction property of equality.
Step-by-step explanation:
For statement 2, we use segment addition postulate to say AC= AE+EC.
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The answer is: " 11 " .
____________________________________________________
→ " x = 11 " .
____________________________________________________
Explanation:
____________________________________________________
Set up the ratio/ proportion as a fraction:
6 cm / 48 cm = 5 cm / (3x + 7) ;
→ The "cm" units cancel out; since: "cm/cm = 1 " ;
→ The "6/48" = "(6 ÷ 6) / (48 ÷ 6) = " 1/8 " ;
→ Rewrite as:
;
Now, we can "cross-multiply" :
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<u>Note</u>:
; 
;
{b
; d } .
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As such:
1 * (3x + 7) = 8 * 5 ;
→ 3x + 7 = 40 ;
Subtract "7" from each side of the equation:
→ 3x + 7 − 7 = 40 <span>− 7 ;
</span>
to get:
→ 3x = 33 ;
Now, divide each side of the equation by "3" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ 3x / 3 = 33 / 3 ;
to get:
____________________________________________________
→ " x = 11 " .
____________________________________________________
→ The answer is: " 11 " .
____________________________________________________
____________________________________________________
→ " x = 11 " .
____________________________________________________
Explanation:
____________________________________________________
Set up the ratio/ proportion as a fraction:
6 cm / 48 cm = 5 cm / (3x + 7) ;
→ The "cm" units cancel out; since: "cm/cm = 1 " ;
→ The "6/48" = "(6 ÷ 6) / (48 ÷ 6) = " 1/8 " ;
→ Rewrite as:
Now, we can "cross-multiply" :
__________________________________________________
<u>Note</u>:
{b
__________________________________________________
As such:
1 * (3x + 7) = 8 * 5 ;
→ 3x + 7 = 40 ;
Subtract "7" from each side of the equation:
→ 3x + 7 − 7 = 40 <span>− 7 ;
</span>
to get:
→ 3x = 33 ;
Now, divide each side of the equation by "3" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ 3x / 3 = 33 / 3 ;
to get:
____________________________________________________
→ " x = 11 " .
____________________________________________________
→ The answer is: " 11 " .
____________________________________________________