In ΔIJK, the measure of ∠K=90°, KJ = 11, IK = 60, and JI = 61. What is the value of the cosine of ∠J to the nearest hundredth?

step-by-step explanation:

12475909

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

30.5*10=350$

you have to multiply the number without the decimal point then add it later but if its tenn u must move the place value of the decimal point 1 to the right.