Answer:
x= 2 5/6
Step-by-step explanation:
1/3= 2/6
2 1/2= 2 3/6
2/6+ 2 3/6= 2 5/6
Hope this helps! :)
Answer:
f(g(x)) = 4x² + 16x + 13
Step-by-step explanation:
Given the composition of functions f(g(x)), for which f(x) = 4x + 5, and g(x) = x² + 4x + 2.
<h3><u>Definitions:</u></h3>
- The <u>polynomial in standard form</u> has terms that are arranged by <em>descending</em> order of degree.
- In the <u>composition of function</u><em> f </em>with function <em>g</em><em>, </em>which is alternatively expressed as <em>f </em>° <em>g,</em> is defined as (<em>f </em> ° <em>g</em>)(x) = f(g(x)).
In evaluating composition of functions, the first step is to evaluate the inner function, g(x). Then, we must use the derived value from g(x) as an input into f(x).
<h3><u>Solution:</u></h3>
Since we are not provided with any input values to evaluate the given composition of functions, we can express the given functions as follows:
f(x) = 4x + 5
g(x) = x² + 4x + 2
f(g(x)) = 4(x² + 4x + 2) + 5
Next, distribute 4 into the parenthesis:
f(g(x)) = 4x² + 16x + 8 + 5
Combine constants:
f(g(x)) = 4x² + 16x + 13
Therefore, f(g(x)) as a polynomial in <em>x</em> that is written in standard form is: 4x² + 16x + 13.
Answer:
C) 6
Step-by-step explanation:
Alternate interior angles are congruent meaning that they are equal to each other sp 18x+2=110 is simplified to 18x=108. Then divide by 18 to get x equal to 6. Hope this helped and post more :)
ince the problem is only asking for 4 years, we can just calculated it out year by year. Recall the formula for compounding interest: A = P(1+r)n, where A is the total amount, P is the principle (amount you start with), r is the interest rate per period of time, and n is the number of periods (in this case, r is annual interest rate, so n is number of years). At the beginning (Year 0), Lou starts off with 10000: A = 10000 At the end of Year 1, Lou earned interest on that amount, plus he has deposited another 5000: A = 10000(1.08) + 5000 End of Year 2, Lou's interest from the year 0 amount has compounded, he has started earning interest on the amount deposited last year, and he deposits another 5000: A = 10000(1.08)2 + 5000(1.08) + 5000 End of Year 3, same idea. Lou has earned compounding interest on all existing deposits, and deposits another 5000: A = 10000(1.08)3 + 5000(1.08)2 + 5000(1.08) + 5000 End of Year 4, same idea: A = 10000(1.08)4 + 5000(1.08)3 + 5000(1.08)2 + 5000(1.08) + 5000 = 36135.45
