The percent increase in enrollment is 6 %
The operation used in first step is finding the difference between final value and initial value
<h3><u>Solution:</u></h3>
Given that This year, 1,272 students enrolled in night courses a a local college
Last year only 1,200 students enrolled.
To find: percent increase in the enrollment
The percent increase between two values is the difference between a final value and an initial value, expressed as a percentage of the initial value.
<em><u>The percent increase is given as:</u></em>
Here initial value (last year) = 1200 and final value(this year) = 1272
Substituting the values in above formula,
Thus percent increase is 6 %
Answer:
$7,925.53
Step-by-step explanation:
We'll have to use the compound interest formula: A = P(1 + r/n)ⁿˣ
A = final amount (?)
P = starting amount (5700)
r = rate
n = times applied (12 since its monthly and there 12 months in a year)
x = years (12)
A = 5700(1 + 0.0275/12)¹⁴⁴
A = 7925.525498629932
<span>The kinetic energy is 6395.03 J</span>
GIVEN :
a = 1/√10 ( 3i + k) and
b = 1/7 ( 2i + 3j - 6k)
TO FIND :
( 2a- b) . [ ( a x b ) x ( a + 2b)]
SOLUTION :
◆Going with the equation given,
( 2a - b) . [ ( a x b ) x ( a + 2b)]
= (2a - b) [( a×b×a) + 2(a×b)×b]
◆BAC - CAB RULE,
A×B×C = B( A.B ) - C(A.B )
= (2a- b ) [ (b (a.a ) - a (a.b ) + 2b ( a.b) -2b (a.b]
Solving further
= (2a - b )(b - 2a)
= -4a.a -b.b
=-5.
Answer:
( 2 - b) . [ ( a x b ) x ( a + 2b)] = -5
Hoped I helped
