Two linear equations can have no solutions, exactly one solution or infinitely many solutions. There will be no solution if the lines are parallel on a graph. There will be exactly one solution if the lines intersect each other on a single point. And finally, there will be infinite solutions if the lines overlap each other perfectly.
A single line however has infinite ordered pair solutions as the line travels infinitely in both directions on the coordinate plane. For example, using the equation y=3x, for any real value of x, we will get a real value for y.
Linear inequalities with two variables have infinitely many solutions. We can use the inequality y>3x as an example. For any real value of x, we will get a real value for y.
I hope this helps!
Step-by-step explanation:
- 4 × (1/7) Less Than
- 12 × 2(5/6) Greater Than
<em>The Rule: </em><em>If</em><em> </em><em>you</em><em> </em><em>are</em><em> </em><em>multiplying</em><em> </em><em>by</em><em> </em><em>a</em><em> </em><em>value</em><em> </em><em>less</em><em> </em><em>than</em><em> </em><em>one</em><em>,</em><em> </em><em>your</em><em> </em><em>product</em><em> </em><em>will</em><em> </em><em>decrease</em><em>.</em><em> </em><em>If</em><em> </em><em>you</em><em> </em><em>are</em><em> </em><em>multiplying</em><em> </em><em>by</em><em> </em><em>1</em><em>,</em><em> </em><em>the</em><em> </em><em>product</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>equal</em><em> </em><em>and</em><em> </em><em>if</em><em> </em><em>you</em><em> </em><em>are</em><em> </em><em>multiplying</em><em> </em><em>by</em><em> </em><em>a</em><em> </em><em>value</em><em> </em><em>greater</em><em> </em><em>than</em><em> </em><em>1</em><em>,</em><em> </em><em>the</em><em> </em><em>product</em><em> </em><em>will</em><em> </em><em>increase</em><em>.</em>
To prove that <span>ΔABC ≅ ΔMQR using SAS, we show that two sides with the intersection angle are congruent.
From the diagram, it is shown that CA is congruent to RM.
From the first option, given that </span>m∠A = 64° and AB = MQ = 31 cm, then we have CA = RM, AB = MQ, and CAB = RMQ (i.e. m∠A = <span>m∠M = 64°). </span>
This shows that the first option is correct.
From the second option, given that CB = MQ = 29 cm, then we have CA = RM, <span>CB = MQ, but ACB is not congruent to RMQ.
Thus the second option in not correct.
From the third option, </span>m∠Q = 56° and CB ≅ RQ, then we have CA = RM, CB = RQ, ACB = 60<span>°, but we do not know the value of MRQ.
Thus the third option is not correct.
From the fourth option, </span>m∠R = 60° and AB ≅ MQ, then we have <span>CA = RM, AB = MQ, RMQ = </span>64<span>°, but we do not know the value of CAB.
Thus the fourth option is not correct.
From the fifth option</span>, <span>AB = QR = 31 cm, then we have </span><span>CA = RM, </span><span>AB = QR, but we do not know the value of CAB or MRQ.
Thus, the fifth option is not correct.
Therefore, the additional information that </span><span>could be used to prove ΔABC ≅ ΔMQR using SAS is </span><span>m∠A = 64° and AB = MQ = 31 cm</span>
Answer:
1/12
Step-by-step explanation:
