Pythagoras Theorem:
hipotenuse²=leg₁²+leg₂²
First posible triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=5 (5²=25)
13³=144 + 25
Answer:can be side lengths of a triangle
Second triangle:
hypotenuse=12.6 (12.6²=158.76)
leg₁=6.7 ( 6.7²=44.89)
leg₂=6.5 (6.5²=42.25)
leg₁²+leg₂²=44.89+42.25=87.14≠158.76
Answer: cannot be side lenghts of a triangle.
third triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=11 (11²=121)
leg₁²+leg₂²=144+121=265≠169
Answer: cannot be side lenghts of a triangle.
fourth triangle:
hypotenuse=13 (13²=169)
leg₁=6 ( 6²=36)
leg₂=4 (4²=16)
leg₁²+leg₁²=36+16=52≠169
Answer: cannot be side lenghts of a triangle.
Answer:
X = 2
Step-by-step explanation:
11 x 2 = 22
22 + 6 = 28
Angles XQL and MQR are congruent because they are vertical angles. So
209 - 13 <em>b</em> = 146 - 4 <em>b</em>
Solve for <em>b</em> :
209 - 13 <em>b</em> = 146 - 4 <em>b</em>
209 - 146 = 13 <em>b</em> - 4<em> b</em>
63 = 9 <em>b</em>
<em>b</em> = 63/9 = 7
Then the measure of angle XQL is
(209 - 13*7)º = 118º
You may find it easiest if you convert all the fractions to decimal form:
5.433...
5.166...
5.756
Now reorder these from smallest to largest.
Note: I had to guess that by "68,90," you actually meant "68/90."
Letter D
A = 135°
B = 30°
cosAcosB - sinAsinB = cos(A+B)
cos(135°+30°)
cos(165°)
