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Naya [18.7K]
4 years ago
13

Two separate bacteria populations that grow each month at different rates are represented by the functions f(x) and g(x). In wha

t month does the f(x) population exceed the g(x) population? Month (x) f(x) = 3x g(x) = 7x + 6 1 3 13 2 9 20 Month 3 Month 4 Month 5 Month 6
Mathematics
2 answers:
Leya [2.2K]4 years ago
8 0

Answer: month 4

Step-by-step explanation

ill try to explain this as simple and short as possible

our equations are

f(x) = 3x   and   g(x) = 7x + 6

we need to see which one grows faster per month. so all you do is plug in the number of months into X in each equation then solve. once you have an answer, plug in the next number of months til you have an answer

we already have the first and second months of each equation so now you just do the next four

for f(x) the growth per month is this

1 month = 3

2 months =9

3 months =27

4 months =81

5 months = 243

and 6 = 729

good, next we for g(x) we plug in the number of months and we get

1 month= 13

2 month = 20

3= 27

4= 34

5= 41

6= 48

SO f(x) starts growing exponentially faster than g(x) during the 4th month, because that's when it starts increasing faster

hope this wasn't too complex, sorry its so long

denis23 [38]4 years ago
3 0

Step-by-step explanation:

This is confusing, what does g(x) equal? There's a bunch of random numbers there. I'll just assume.

They stated that f(x) needs to exceed g(x), so make an inequality plugging in the values

3x > 7x + 6

You see, function f(x) needs to be higher than function g(x)

Subtract 7x from both sides to get:

- 4x > 6

Divide both sides by -4, when you divide by a negative number the signs switch so it is now less than.

x < -1.5

And obviously that's not an answer up there, so you made a mistake stating the values of the functions

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Anyone can do this please if you can do all of it of you can I'm really bad at math​
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Rosemary walks each week for exercise. Let d represent the distance walked and h represent the number of hours spent walking.
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The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg
julia-pushkina [17]

Answer:

Approximately 101\; \rm ft (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

  • Let \rm O denote the observer.
  • Let \rm A denote the top of the tree.
  • Let \rm R denote the base of the tree.
  • Let \rm B denote the point where line \rm AR (a vertical line) and the horizontal line going through \rm O meets. \angle \rm B\hat{A}R = 90^\circ.

Angles:

  • Angle of elevation of the base of the tree as it appears to the observer: \angle \rm B\hat{O}R = 51^\circ.
  • Angle of elevation of the top of the tree as it appears to the observer: \angle \rm B\hat{O}A = 57^\circ.

Let the length of segment \rm BR (vertical distance between the base of the tree and the base of the hill) be x\; \rm ft.

The question is asking for the length of segment \rm AB. Notice that the length of this segment is \mathrm{AB} = (x + 20)\; \rm ft.

The length of segment \rm OB could be represented in two ways:

  • In right triangle \rm \triangle OBR as the side adjacent to \angle \rm B\hat{O}R = 51^\circ.
  • In right triangle \rm \triangle OBA as the side adjacent to \angle \rm B\hat{O}A = 57^\circ.

For example, in right triangle \rm \triangle OBR, the length of the side opposite to \angle \rm B\hat{O}R = 51^\circ is segment \rm BR. The length of that segment is x\; \rm ft.

\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}.

Rearrange to find an expression for the length of \rm OB (in \rm ft) in terms of x:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}.

Similarly, in right triangle \rm \triangle OBA:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}.

Equate the right-hand side of these two equations:

0.810\, x \approx 0.649\, (x + 20).

Solve for x:

x \approx 81\; \rm ft.

Hence, the height of the top of this tree relative to the base of the hill would be (x + 20)\; {\rm ft}\approx 101\; \rm ft.

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