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3 years ago

PLS HELP / EXPLAIN !!!!

1 answer:

5 0

Area of a triangle = $\frac{1}{2}$ · base x height
   180 = $\frac{1}{2}$ · base x 40
   180 = 20 · base
      9 = base

Pythagorean Theorem:
base² + height² = hypotenuse²
9²    + 40² = hypotenuse²
81 + 1600 = hypotenuse²
1681    = hypotenuse²
   √1681           = √hypotenuse²
   41    = hypotenuse

Answer: the length of the hypotenuse is 41 m

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Solve the following ODE's: c) y* - 9y' + 18y = t^2

Nastasia [14]

Answer:

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = $C_1e^{3t}+C_2e^{6t}$

now calculating P.I.

$P.I. = \frac{t^2}{D^2 - 9D +18}$

$P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

hence the complete solution

y = C.F. + P.I.

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

7 0

3 years ago

A figure is made of 2 rectangular prisms...<br> what is the volume?

MaRussiya [10]

The volume of the figure that is composed of two rectangular prisms is: 146 cubic in.

<h3>What is the Volume of a Rectangular Prism?</h3>

The volume of a rectangular prism = (length)(width)(height).

The figure can be decomposed into two rectangular prisms. Find the volume of each of the prisms.

Volume of rectangular prism 1:

Length of the prism = 8 in.

Width of the prism = 7 in.

Height of the prism = 1 in.

Volume of rectangular prism 1 = (8)(7)(1) = 56 in.³

Volume of rectangular prism 2:

Length of the prism = 10 in.

Width of the prism = 9 in.

Height of the prism = 1 in.

Volume of rectangular prism 1 = (10)(9)(1) = 90 in.³

Volume of the figure = 90 + 56 = 146 cubic in.

Learn more about volume of rectangular prism on:

brainly.com/question/12917973

#SPJ1

7 0

2 years ago

In how many ways can a teacher arrange 6 students in the front row of a classroom with a total of 20 students?

Luba_88 [7]

The concept of the question above is Permutation where in we are to arrange to arrange the 6 students from which there are 20 students. This is the "permutation of 20 taken 6" or 20P6. The numerical value of the permutation is equal to 27907200.

5 0

3 years ago

An office has 150 employees, and 30 of the employees work the night shift. What percentage of the employees work the night shift

Zanzabum

<h2>Answer:</h2>

The percent of employees who work in the night shift are:

20%

<h2>Step-by-step explanation:</h2>

Total number of employee in the office are: 150

Number of employee who work in the night shift are: 30

The percent of the employee who work in the night shift is calculated as follows:

$\dfrac{\text{Number\ of\ people\ working\ in\ the\ night\ shift}}{\text{Total\ number\ of\ employees}}\times 100$

Hence, on putting the value in the above formula we get:

$=\dfrac{30}{150}\times 100\\\\\\=\dfrac{1}{5}\times 100\\\\\\=20\%$

Hence, the answer is:

20%

4 0

4 years ago

Read 2 more answers

Of interest is to test the hypothesis that the mean length of all face-to-face meetings and the mean length of all Zoom meetings

Goshia [24]

Answer:

Null hypothesis: $\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n$

Alternative hypothesis: $\mu_i \neq \mu_j , i,j =1,2,....,n$

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

Step-by-step explanation:

For this case we wnat to test if all the mean length of all face-to-face meetings and the mean length of all Zoom meetings are the same. So then the system of hypothesis are:

Null hypothesis: $\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n$

Alternative hypothesis: $\mu_i \neq \mu_j , i,j =1,2,....,n$

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

6 0

3 years ago