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 %
About 101.44 grams per liter
Mark brainliest please
Hope this helps you
Answer:
4,1
Step-by-step explanation:
(x-4)(x-1)
x=4 or x=1
See attachment for full explanation.
The polynomial p(x)=x^3+7x^2-36p(x)=x 3 +7x 2 −36p, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 7, x, square
Iteru [2.4K]
Answer:
(x-2)(x+3)(x+6)
Step-by-step explanation:
Given the polynomial function p(x)=x^3+7x^2-36
We are to write it as a product of its linear factor
Assuming the value of x that will make the polynomial p(x) to be zero
Let x = 2
P(2) = 2³+7(2)²-36
P(2) = 8+7(4)-36
P(2) = 8+28-36
P(2) = 0
Since p(2) = 0 hence x-2 is one of the linear factors
Also assume x = -3
P(-3) = (-3)³+7(-3)²-36
P(-3) = -27+7(9)-36
P(-3) = -27+63-36
P(-3) = 36-36
P(-3) = 0
Since p(-3) = 0, hence x+3 is also a factor
The two linear pair are (x-2)(x+3)
(x-2)(x+3) = x²+3x-2x-6
(x-2)(x+3) = x²+x-6
To get the third linear function, we will divide x^3+7x^2-36 by x²+x-6 as shown in the attachment.
x^3+7x^2-36/x²+x-6 = x+6
Hence the third linear factor is x+6
x^3+7x^2-36 = (x-2)(x+3)(x+6)
