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aev [14]
4 years ago
14

A town has a population of 1400 and grows at 3% every year. What will be the population after 6 years, to the nearest whole numb

er?
Mathematics
1 answer:
crimeas [40]4 years ago
7 0

Answer:1672

Step-by-step explanation:add 3% from each previous year to get to the 6th year

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Obtuse scalene triangle

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Design a fish tank in the shape of a rectangular prism that has a length of 24ft and has volume of 2600ft^3. What piece of infor
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The cost to a store for a box of cereal is $2.50. The store is selling the box of cereal for $3.50. What is the percent of marku
zysi [14]
Well, the cereal cost 2.5, and the store is selling for 3.5, so is 1 buck more.

if we take 2.5 to be the 100%, what is 1 off of it in percentage?

\bf \begin{array}{ccll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
2.5&100\\
1&p
\end{array}\implies \cfrac{2.5}{1}=\cfrac{100}{p}\implies p=\cfrac{1\cdot 100}{2.5}\implies p=\stackrel{\%}{40}
8 0
4 years ago
PLZ HELP, DUE TONIGHT!! Find the values of a and b such that f(x) is continuous at x=
Goryan [66]

Answers:

a = 2 and b = -4

============================================================

Explanation:

Let's define the three helper functions

  • f(x) = ax^2 - b
  • h(x) = 6
  • j(x) = 5ax+b

which are drawn from the piecewise function. The g(x) function will change depending on what the input is.

  • If x < 1, then g(x) = f(x).
  • If x = 1, then g(x) = h(x)
  • If x > 1, then g(x) = j(x)

Since we want g(x) to be continuous at x = 1, this must mean the three functions f(x), h(x), j(x) must have the same output value when the input is x = 1.

Because h(x) = 6 is a constant function, the output is always 6 regardless of the input. Therefore, we want f(x) and j(x) to have 6 as their output when x = 1. Or else, the pieces won't connect.

Plug x = 1 into the f(x) function to get

f(x) = ax^2 - b

f(1) = a(1)^2 - b

f(1) = a - b

Set this equal to the desired output of 6 and we end up with the equation a-b = 6. Solving for 'a' leads to a = b+6.

------------

We'll use the same idea for j(x)

j(x) = 5ax + b

j(1) = 5a(1) + b

j(1) = 5a + b

5a+b = 6

5(b+6) + b = 6 ... plug in a = b+6; solve for b

5b+30+b = 6

6b+30 = 6

6b = 6-30

6b = -24

b = -24/6

b = -4

Which then leads to,

a = b+6

a = -4+6

a = 2

------------

Since a = 2 and b = -4, we go from this

g(x) = \begin{cases}ax^2-b, \ \ x < 1\\6, \ \ x = 1\\5ax+b, \ \ x > 1\end{cases}

to this

g(x) = \begin{cases}2x^2+4, \ \ x < 1\\6, \ \ x = 1\\10x-4, \ \ x > 1\end{cases}

Meaning

f(x) = 2x^2+4 and j(x) = 10x-4

You should find that plugging x = 1 into each of those two functions leads to 6 as the output.

The graph is shown below. Note the red graph f(x) is only drawn when x < 1. Similarly, j(x) is only drawn when x > 1. The orange point represents h(x) which only happens when x = 1. So as the name implies, the piecewise function g(x) is composed of pieces of the three functions f(x), h(x), j(x).

3 0
3 years ago
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