Find percent increase by subtracting the amounts and dividing it by the old amount, then multiply the result by 100 to make it a percent.
Increase= (277-210)/210
67/210
Then it ends up as 0.319047619...
Multiply that by 100 to make it a percent, which is..
31.9047619..
Round it off, and you get..
32%!
First off, see how much 8700 as principal, yields at 3% APR
that is
it will yield some amount
subtract that amount from 393
the difference is how much the yield will be on the 6% investment
so
Hello!
For #1, the LCD is (A) 12 since both numbers go into 12 as their minimal number.
For #2, the missing number is (B) 4 because 1/4*4/4=4/16.
For #3, the answer is (B) 3 because you have to crossmultiply 20n(4*15) to find the answer.
for #4, the correct answer is (A) 4/5 is less than 5/6 because when converted into fractions with a common denominator, 24/30 is less than 25/30.
For #5, the LCD is 36; therefore, the answer is (A) because the fractions in the answer are properly converted to their new denominator.
I really hope I helped!
Answer:
Direct Variation
Step-by-step explanation:
The relationship between two variables such that y = kx if k is a nonzero number. Also, as one quantity increases, the second quantity increases or as one quantity decreases, the second quantity decreases. Therefore ac=5 is a direct variation
Assuming P (usually written in upper case) represents a force normal to a given cross section.
If a point load is applied to any point of the section, stress concentration will cause axial stress to vary.
The context of the question considers the uniformity of axial stress at a certain distance away from the point of application (thus stress concentration can be neglected).
If a force P is applied through the centroid, sections will be stressed uniformly. However, if the force P is applied at a distance "e" from the centroid, the equivalent load on the section equals an axial force and a moment Pe. The latter causes bending of the member, causing non-uniform stress.
If we assume A=(uniform) cross sectional area, and I=moment of inertia of the section, then stress varies with the distance y from the centroid equal to
stress=sigma=P/A + My/I
where P=axial force, M=moment = Pe.
Therefore when e>0, the stress varies across the section.
