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3 years ago

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a science museum offers tours for school groups with 20 or more students. the tour costs $100 plus $3.50 for every student in ex

cess of 20 students. a school group pays $345 for the tour. part A: write an equation that can be used to determine the number of students,n, who went in the tour. explain how toy determined your answer

2 answers:

Svetllana [295]3 years ago

7 0

Answer:

Equation : 100 + (n - 20) × 3.5 = 345

90 students went for tour.

Step-by-step explanation:

Since, the cost of first 20 students is fix and it is $100

and, each student after these 20 students has to pay $3.50, also tour cost is $345.

Let the number of total student went for tour be "n".

Hence, we get equation,

100 + (n - 20) × 3.5 = 345

⇒ (n - 20) × 3.5 = 345 - 100

⇒ (n - 20) × 3.5 = 245

⇒ (n - 20) = 245 ÷ 3.5

⇒ (n - 20) = 70

⇒ n = 90 students

ohaa [14]3 years ago

5 0

The total paid was $345. Twenty students cost $100.00. $345-4100=$245
245/3.50=70 additional students. 70+20=90 students went on the tour

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An arrow is shot vertically upward at a rate of 240 feet per second from ground level. Use the projectile formula to determine w

Juli2301 [7.4K]

Answer:

After 14.92 seconds, the arrow hits the ground.

Step-by-step explanation:

Given:

Initial velocity of the arrow is, $u=240\ ft/s or 73.152\ m/s$

The arrow goes up and come back to the same starting position. So, initial position is same as that of final position.

Therefore, displacement of the arrow is 0 m.

The displacement of the arrow is given as:

$s=ut-\frac{1}{2}gt^{2}$ (equation 1)

Plug all value in equation 1 and solve for $t$

$0=73.152\timests t-\frac{1}{2} \times9.8 \times t^{2}$

$0=73.152\timests t-4.9 \times t^{2}$

$0=t(73.152-4.9 \times t)$

$73.152-4.9\times t=0$ where $t\neq 0$

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3 years ago

Use Gaussian elimination to write each system in triangular form

Feliz [49]

Answer:

To see the steps to the diagonal form see the step-by-step explanation. The solution to the system is $x = -\frac{1}{9}$ , $y= -\frac{1}{9}$ , $z= \frac{4}{9}$ and $w = \frac{7}{9}$

Step-by-step explanation:

Gauss elimination method consists in reducing the matrix to a upper triangular one by using three different types of row operations (this is why the method is also called row reduction method). The three elementary row operations are:

Swapping two rows
Multiplying a row by a nonzero number
Adding a multiple of one row to another row

To solve the system using the Gauss elimination method we need to write the augmented matrix of the system. For the given system, this matrix is:

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

For this matrix we need to perform the following row operations:

$R_2 - 1 R_1 \rightarrow R_2$ (multiply 1 row by 1 and subtract it from 2 row)
$R_3 + 1 R_1 \rightarrow R_3$ (multiply 1 row by 1 and add it to 3 row)
$R_4 - 1 R_1 \rightarrow R_4$ (multiply 1 row by 1 and subtract it from 4 row)

$R_2 \leftrightarrow R_3$ (interchange the 2 and 3 rows)

$R_2 / 2 \rightarrow R_2$ (divide the 2 row by 2)
$R_1 - 1 R_2 \rightarrow R_1$ (multiply 2 row by 1 and subtract it from 1 row)
$R_4 - 1 R_2 \rightarrow R_4$ (multiply 2 row by 1 and subtract it from 4 row)
$R_3 \cdot ( -1) \rightarrow R_3$ (multiply the 3 row by -1)
$R_2 - 1 R_3 \rightarrow R_2$ (multiply 3 row by 1 and subtract it from 2 row)
$R_4 + 3 R_3 \rightarrow R_4$ (multiply 3 row by 3 and add it to 4 row)
$R_4 / 4.5 \rightarrow R_4$ (divide the 4 row by 4.5)

After this step, the system has an upper triangular form

The triangular matrix looks like:

$\left[\begin{array}{cccc|c}1 & 0 & 0 & -0.5 & -0.5 \\0 & 1 & 0 & -0.5 & -0.5\\0 & 0 & 1 & 2 & 2 \\0 & 0 & 0 & 1 & \frac{7}{9}\end{array}\right]$

If you later perform the following operations you can find the solution to the system.

$R_1 + 0.5 R_4 \rightarrow R_1$ (multiply 4 row by 0.5 and add it to 1 row)
$R_2 + 0.5 R_4 \rightarrow R_2$ (multiply 4 row by 0.5 and add it to 2 row)
$R_3 - 2 R_4 \rightarrow R_3$ (multiply 4 row by 2 and subtract it from 3 row)

After this operations, the matrix should look like:

$\left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & -\frac{1}{9} \\0 & 1 & 0 & 0 & -\frac{1}{9}\\0 & 0 & 1 & 0 & \frac{4}{9} \\0 & 0 & 0 & 1 & \frac{7}{9}\end{array}\right]$

Thus, the solution is:

$x = -\frac{1}{9}$ , $y= -\frac{1}{9}$ , $z= \frac{4}{9}$ and $w = \frac{7}{9}$

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<h3>Answer: Choice B</h3>

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

Further explanation:

Let's go through the answer choices one by one

A. This is false because 0 does not come before -5, but instead -5 is listed first. The order -5,-4,-3,-2,-1,0 is correct meaning that $-5 \le x \le 0$ is the correct order as well.
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