Answer:
The triangle has two sides that are equivalent and the angle where the two meet are also equivalent.
In other words, two sides and the angle between them are congruent.
Hope this helps, if it does please give me brainliest, it will help me a lot :)
Have a good day
Hello,
Very nice as problem.
2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies
since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0
The easy part is isolating the absolute-value term:
5 + 7 |2<em>x</em> - 1| = -44
7 |2<em>x</em> - 1| = -49
|2<em>x</em> - 1| = -7
Remember that the absolute value function returns a positive number that you can think of as the "size" of that number, or the positive distance between that number and zero. If <em>x</em> is a positive number, its absolute value is the same number, |<em>x</em>| = <em>x</em>. But if <em>x</em> is negative, then the absolute value returns its negative, |<em>x</em>| = -<em>x</em>, which makes it positive. (If <em>x</em> = 0, you can use either result, since -0 is still 0.)
The important thing to take from this is that there are 2 cases to consider: is the expression in the absolute value positive, or is it negative?
• If 2<em>x</em> - 1 > 0, then |2<em>x</em> - 1| = 2<em>x</em> - 1. Then the equation becomes
2<em>x</em> - 1 = -7
2<em>x</em> = -6
<em>x</em> = -3
• If 2<em>x</em> - 1 < 0, then |2<em>x</em> - 1| = - (2<em>x</em> - 1) = 1 - 2<em>x</em>. Then
1 - 2<em>x</em> = -7
-2<em>x</em> = -8
<em>x</em> = 4
