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3 years ago

Find the exact length of the curve. x = 9 + 9t2, y = 6 + 6t3, 0 ≤ t ≤ 4

Mathematics

1 answer:

andrew11 [14]3 years ago

3 0

ANSWER

$=102 \sqrt{ 17}$

EXPLANATION

The equation of the curve is:

$x = 9 + 9 {t}^{2}$

$y = 6 + 6 {t}^{3}$

We differentiate to obtain:

$\frac{dx}{dt} = 18t$

$\frac{dy}{dt} = 18t$

The length of the arc

$l = \int _0 ^{4} \sqrt{ {( \frac{dx}{dt} )}^{2} + {( \frac{dy}{dt} )}^{2} } dt$

This implies that:

$= \int _0 ^{4} \sqrt{ {( 18t )}^{2} + {( {18 {t}^{2} } )}^{2} } dt$

$= \int _0 ^{4} 18t \sqrt{ 1+ { {18 {t}^{2} } }} dt$

$=102 \sqrt{ 17}$

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The sum of the interior angle of an octagon is

ehidna [41]

Answer:

1080 degrees

Step-by-step explanation:

A regular octagon has 135 degrees per interior angle. To find the sum, add 135 for each angle. There are 8 interior angles.

8*135 = 1,080

8 0

3 years ago

Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for th

Masteriza [31]

Answer:

Let $t_i$ be the time for the $i$ th task.

We know these times have a certain structure:

Any 3 adjacent tasks will take half as long as the next two tasks.

In the form of an equations we have

$t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5 \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6$

The second task takes 1 second $t_2=1$

The fourth task takes 10 seconds $t_4=10$

So, we have the following system of equations:

$t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0 \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10$

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

$\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

Subtract row 2 from row 1 $\left(R_1=R_1-R_2\right)$ .
Subtract row 2 from row 3 $\left(R_3=R_3-R_2\right)$ .
Add row 3 to row 2 $\left(R_2=R_2+R_3\right)$ .
Multiply row 3 by −1 $\left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right)$ .
Add row 4 multiplied by $\frac{3}{2}$ to row 1 $\left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right)$ .
Subtract row 4 from row 3 $\left(R_3=R_3-R_4\right)$ .

Here is the reduced echelon form for the augmented matrix.

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

c) The additional rows are

$\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right$

and the augmented matrix is

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]$

d) To solve the system you must use these row operations.

Subtract row 1 from row 6 $\left(R_6=R_6-R_1\right)$ .
Subtract row 2 from row 6 $\left(R_6=R_6-R_2\right)$ .
Subtract row 3 from row 6 $\left(R_6=R_6-R_3\right)$ .
Swap rows 5 and 6.
Add row 5 to row 3 $\left(R_3=R_3+R_5\right)$ .
Multiply row 5 by 2 $\left(R_5=\left(2\right)R_5\right)$ .
Subtract row 6 multiplied by 1/2 from row 1 $\left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right)$ .
Add row 6 multiplied by 1/2 to row 3 $\left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right)$ .

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]$

The solutions are: $(t_1,...,t_6)=(5,1,44,10,90,20)$ .

5 0

3 years ago

A toy cannon ball is launched from a cannon on top of a platform. The equation h(t) =- 5<img src="https://tex.z-dn.net/?f=t%5E%7

DanielleElmas [232]

Answer:

Part A)

Part B)

About 2.9362 seconds.

Step-by-step explanation:

The equation $\displaystyle h(t)=-5t^2+14t+2$ models the height h in meters of the ball t seconds after its launch.

Part A)

To determine whether or not the ball reaches a height of 14 meters, we can find the vertex of our function.

Remember that the vertex marks the maximum value of the quadratic (since our quadratic curves down).

If our vertex is greater than 14, then, at some time t, the ball will definitely reach a height of 14 meters.

However, if our vertex is less than 14, then the ball doesn’t reach a height of 14 meters since it can’t go higher than the vertex.

So, let’s find our vertex. The formula for vertex is given by:

$\displaystyle (-\frac{b}{2a},h(-\frac{b}{2a}))$

Our quadratic is:

$\displaystyle h(t)=-5t^2+14t+2$

Hence: a=-5, b=14, and c=2.

Therefore, the x-coordinate of our vertex is:

$\displaystyle x=-\frac{14}{2(-5)}=\frac{14}{10}=\frac{7}{5}$

To find the y-coordinate and the maximum height, we will substitute this value back in for x and evaluate. Hence:

$\displaystyle h(\frac{7}{5})=-5(\frac{7}{5})^2+14(\frac{7}{5})+2$

Evaluate:

$\displaystyle \begin{aligned} h(\frac{7}{2})&=-5(\frac{49}{25})+\frac{98}{5}+2 \\ &=\frac{-245}{25}+\frac{98}{5}+2\\ &=\frac{-245}{25}+\frac{490}{25}+\frac{50}{25}\\&=\frac{-245+490+50}{25}\\&=\frac{295}{25}=\frac{59}{5}=11.8\end{aligned}$

So, our maximum value is 11.8 meters.

Therefore, the ball doesn’t reach a height of 14 meters.

Part B)

To find out how long the ball is in the air, we can simply solve for our t when h=0.

When the ball stops being in the air, this will be the point at which it is at the ground. So, h=0. Therefore:

$0=-5t^2+14t+2$

A quick check of factors will reveal that is it not factorable. Hence, we can use the quadratic formula:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Again, a=-5, b=14, and c=2. Substitute appropriately:

$\displaystyle x=\frac{-(14)\pm\sqrt{(14)^2-4(-5)(2)}}{2(-5)}$

Evaluate:

$\displaystyle x=\frac{-14\pm\sqrt{236}}{-10}$

We can factor the square root:

$\sqrt{236}=\sqrt{4}\cdot\sqrt{59}=2\sqrt{59}$

Hence:

$\displaystyle x=\frac{-14\pm2\sqrt{59}}{-10}$

Divide everything by -2:

$\displaystyle x=\frac{7\pm\sqrt{59}}{5}$

Hence, our two solutions are:

$\displaystyle x=\frac{7+\sqrt{59}}{5}\approx2.9362\text{ or } x=\frac{7-\sqrt{59}}{5}\approx-0.1362$

Since our variable indicates time, we can reject the negative solution since time cannot be negative.

Hence, our zero is approximately 2.9362.

Therefore, the ball is in the air for approximately 2.9362 seconds.

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3 years ago

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goblinko [34]

Alright, here we go ;)

10000=5000*1.07^10.2

40000=5000*1.07^x

Systems of equations and solve.

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3 years ago

How to find the factor of a trinomial?

BaLLatris [955]

I can tell you about the 'ac' method of factoring trinomials by way of an example:-

Factor 3x^2 - 4x - 15

first multiply the first number by the last That is 3 * -15 = -45

Now we need to find 2 factors of -45 which will when added give - 4 ( the coefficient of x)

-9 and = 5 look like the ones so we write

3x^2 - 9x + 5x - 15

Now factor this by grouping:-

= 3x(x - 3) + 5(x - 3)

x - 3 is common to the 2 parts so we have:-

(3x + 5)(x - 3) <-------- These are your factors

Any trinomial can be factored in this way.

6 0

4 years ago