How you do this please help me due tomorrow

To solve this problem we must propose an equation that models the number of kilometers of bicycle trails made monthly in Charles City and in Tinsel Town

For Charles City we know that the number of kilometers built in the first month is 5, and it doubles every month. Then we have an exponential equation, in base 2, whose initial value is 5.

This equation has the following form:

$y = C_1 (2^{x-1})$

Where:

$C_1$ is the number of kilometers built in month 1

x is the number of months: {1, 2, 3, 4, 5 ... x}

x = 1 represents month 1.

So:

$Charles\ City\ (km) = 5(2^{x-1})$

For Tinsel Town we know that the number of kilometers built in the first month is 21 and increases at a fixed rate of 5 kilometers per month. This can be modeled by a linear equation.

$Tinsel\ Town\ (km) = 21 + 5(x-1)$

Where x is the number of months. x: {1, 2, 3, 4, 5 ...}

We want to know at the end of what month the total length of the cycle lanes in Charles City first exceeds the length in Tinsel Town

Then we equal both equations and clear x.

$5 (2^{x-1}) = 21 + 5(x-1)\\\\5 (2^{x-1}) - (21 + 5(x-1)) = 0$

Clearing x from this equation is very difficult, so to find x, we iterate until we get the value of x that causes the equation to approach 0.

For x = 3

$5(2^{3-1}) - (21 +5(3-1)) = 0\\\\20- 31= -11$

For x = 4

$5(2^{4-1}) - (21 +5(4-1)) = 0\\\\40 - 36 = 4$

The value must be between x = 3 and x = 4

For x = 3.8

$5 (2^{3.8-1}) - (21 +5(3.8-1)) = 0\\\\34.82 - 35 = 0.18$

Then x ≈ 3.8 months.

Finally we have that By the end of the fourth month the total length of the cycle lanes in Charles City exceeds the length in Tinsel Town

6 0

4 years ago

21 Maria sold small boxes of candy for $1 and

Anna35 [415]

Answer:

There where 10 more Small boxes sold than Large boxes.

Step-by-step explanation:

This is a typical question that can be solved by obtaining a system of two linear equations and solving for each variable.

In principle Linear Equations are algebraic expressions denoting a relationship between a Dependent variable $y$ and an Independent variable $x$ . In a system of Two Linear equations we have two equations of the same variable sets (thus Two Independent variables) so in this case both $y$ and $x$ will be variable terms.

Now with respect to the question and the given information, here our two Variable terms will be the small and the large boxes.

Given Information:

Small Boxes (lets call them $s$ ) cost $1 per box

Large Boxes (lets call them $l$ ) cost $4 per box
Total Number of Boxes sold is 30
Total Profit from sold Boxes is $60

Thus from the above we can obtain one equation denoting the Total Number of Boxes sold and one equation denoting the Total Profit from sold boxes, respectively, as follow:

$s+l=30$ Eqn(1): Total Number of Boxes

$1s+4l=60$ Eqn(2): Total Profit from sold boxes

Now we have a system of two linear equations which we can solve and find the number of small and large boxes, $s$ and $l$ respectively.

From Eqn(1) we see that

$s=30-l$ Eqn(3).

Plugging Eqn(3) in Eqn(2) we can solve for $l$ as:

$1(30-l)+4l=60$

$30-l+4l=60$ Factored out bracket

$3l=60-30$ Gather similar terms on each side and simplify

$3l=30$

$l=\frac{30}{3}$

$l=10$

Plugging in the value for $l=10$ in Eqn(3) we have

$s=30-10\\s=20$

So we know that they were 10 Large Boxes and 20 Small Boxes sold, thus to answer our question, there where 10 more Small boxes sold than Large boxes.

5 0

4 years ago

Question for you.How much a day do you use your car a day?(For project)

grin007 [14]

I don't use a car because I'm only thirteen but, if consider a bus a car I ride the bus in the mornings to school and I am a car rider on the way home

8 0

3 years ago