Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides). If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.
Let's see whether they do.
8^2 + 8^2 = 11^2 ???
64 + 64 = 121? NO. This is not a right triangle.
If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:
8^2 + 8^2 = d^2, where d=hypotenuse. Then 64+64 = d^2, and
d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.
11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.
The answer to your question would be C 6 can be put into 18 and 24
Answer:
a.) x = 5.36
b.) y = - 0.001
c.) t = 11.26
d.) k = 4.57
Step-by-step explanation:
a) x−1.27=4.09
b) 2.3−y=2.301
c) t+3.04=14.3
d) 1.1+k=5.67
Solving. right?
a.) x = 4.09 + 1.27 = 5.36
b.) y = 2.3 - 2.301 = - 0.001
c.) t = 14.3 - 3.04 = 11.26
d.) k = 5.67 - 1.1 = 4.57
Answer:
i think its set B..........
