Remember a^(m) x a^(n) = a(m+n) & √a = a^(1/2)
===> x^(1/2 -3/2)+x(1/2 -1/2) ====> x^(-1)+x^(0) (any number Exp 0 =1)
X^(-1) +1 ===> 1/x +1 or (1+x)/x
Let the two ages be j and m, respectively. Then j+m>32.
Solving for Mary's age, we get m > 32 - j. Because j = m + 2, m > 32 - (m+2).
Continue solving for m: Adding m to both sides of this inequality results in
2m > 32 - 2. Then 2m > 30, and m > 15. Mary's age is greater than 15.
XZ ≅ EG and YZ ≅ FG is enough to make triangles to be congruent by HL. Option b is correct.
Two triangles ΔXYZ and ΔEFG, are given with Y and F are right angles.
Condition to be determined that proves triangles to be congruent by HL.
<h3>What is HL of triangle?</h3>
HL implies the hypotenuse and leg pair of the right-angle triangle.
Here, two right-angle triangles ΔXYZ and ΔEFG are congruent by HL only if their hypotenuse and one leg are equal, i.e. XZ ≅ EG and YZ ≅ FG respectively.
Thus, XZ ≅ EG and YZ ≅ FG are enough to make triangles congruent by HL.
Learn more about HL here:
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In ΔXYZ and ΔEFG, angles Y and F are right angles. Which set of congruence criteria would be enough to establish that the two triangles are congruent by HL?
A.
XZ ≅ EG and ∠X ≅ ∠E
B.
XZ ≅ EG and YZ ≅ FG
C.
XZ ≅ FG and ∠X ≅ ∠E
D.
XY ≅ EF and YZ ≅ FG
7)X=20
8)X=14
9)X=45.83 repeating
10)X=-19
11)X=100
12)X=-25
Answer:
y=a(b)^x
y=0.01(3)^x
a is start, b is multiplier
