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The function, B (x), models the estimated tuition cost where x is the number

zepelin [54]

The expression that completes the function b(x) is b(x) = 33741 * (1.028)^x

<h3>How to determine the expression of b(x)?</h3>

The given parameters are:

Initial value, a = 33741

Rate, r = 2.8%

The cost of tuition each year since 2015 is represented as

B(x) = a * (1 + r)^x

This gives

B(x) = 33741 * (1 + 2.8%)^x

Evaluate

b(x) = 33741 * (1.028)^x

Hence, the expression that completes the function b(x) is b(x) = 33741 * (1.028)^x

Read more about exponential functions at:

brainly.com/question/12636638

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Complete question

A study estimates that the cost of tuition at a university will increase by 2.8% each year. The cost of tuition at the University in 2015 was $33,741 the function b(x) , models the estimated tuition cost , where x is the number of years since 2015.

Find the expression that completes the function b(x)

4 0

2 years ago

Drinking 8 fluid ounces of milk provides 270.0 milligrams of calcium. How many fluid ounces of milk provide 72.5 milligrams of c

Margarita [4]

Set up a proportion:

$\frac{8}{270} = \frac{x}{72.5}$

The numerator represents fluid ounces of milk, and the denominator represents milligrams of calcium.

Simplify the first fraction to make multiplication easier:

$\frac{8}{270} \div \frac{2}{2} = \frac{4}{135}$
$\frac{4}{135} = \frac{x}{72.5}$

Cross multiply the fractions:

$72.5 * 4 = 290$
$135x = 290$

Divide both sides by 135 to get x by itself:

$290 \div 135 = 2.1$
$x = 2.1$

2.1 fluid ounces of milk provide 72.5 milligrams of calcium.

3 0

3 years ago

Read 2 more answers

Please help need this before 4:15pm

yawa3891 [41]

Answer:

Top right i believe

Step-by-step explanation: the elements that make up life are carbon, hydrogen, oxygen, nitrogen, phosphorus, and sulfur.

7 0

3 years ago

What is the sum of -1/11 and 5/11 4/11 6/11 -4/11 -6/11

kaheart [24]

-1/11 + 5/11
= 5/11 - 1/11
= 4/11

Answer
4/11

6 0

3 years ago

Read 2 more answers

Can someone explain to me why are we adding 2kpi when we are doing zeros for sin and cos, but adding kpi when doing zeros for tg

makvit [3.9K]

on the first exercise, you got a solution angle of π/18, that's a good solution for the I Quadrant only, however, on a circle, we have angles that go from 0 to 2π, however we can always keep on going around and continute to 2π + π/2 or 3π or 4π, or 115π/3 or 1,000,000π/18 and so on, and we're really just going around the circle many times over, getting a larger and larger angle, same circular motion.

π/18 on that exercise works for the I Quadrant, however if we continue and go around say 2π, we'll find that 2π/3 + π/18 is a coterminal angle with π/18, and thus that angle has also the same sine value.

π/18 + 2kπ/3 , where k = integer, is a way to say, all angles around the circle that look like this have the same sine, namely

π/18 + 2(1)π/3

π/18 + 2(2)π/3

π/18 + 2(3)π/3

π/18 + 2(5)π/3

π/18 + 2(99999999)π/3

.....

so using the "k" as some sequence multiplier, is a generic notational way to say, "all these angles".

you'll also find that "n" is used as well for the same notation, say for example

2π/3 + 2πn.

8 0

3 years ago