<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B10%7D%20%20%3D%20%20%5Cfrac%7B1%7D%7B5%7D%20" id="TexFormula1" title=" \fr

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VashaNatasha [74]

3 years ago

ac{x}{10} = \frac{1}{5} " alt=" \frac{x}{10} = \frac{1}{5} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

4 0

You may solve this ratio by cross multiplying. To do this, you multiply the numbers diagonally across from each other.

x/10 = 1/5
10(1) = x(5)
10 = 5x
2 = x

Answer:
x = 2

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Need help with this please dont cheat and put just b#l1$H!t

DIA [1.3K]

Answer:

B (ignore this its for 20 characters

4 0

3 years ago

6. A super-bouncy-ball is thrown 25m into the air. The ball falls, rebounds to 70% of the height of the previous bounce and fall

inn [45]

Answer:

The total distance the ball has travelled after it hits the ground the 5th time is of 48.53m.

Step-by-step explanation:

Geometric sequence:

In a geometric sequence, the quotient of consecutive terms is the same. The nth term of a geometric sequence is given by:

$a_n = a_1q^{n-1}$

In which $a_1$ is the first term and $q$ is the common ratio.

A super-bouncy-ball is thrown 25m into the air. The ball falls, rebounds to 70% of the height of the previous bounce and falls again.

This means that:

$q = 0.7, a_1 = 25*0.7 = 17.5$

Then

$a_n = a_1q^{n-1}$

$a_n = 17.5(0.7)^{n-1}$

First 5 terms:

$a_1 = 17.5$

$a_2 = 17.5(0.7)^{2-1} = 12.25$

$a_3 = 17.5(0.7)^{3-1} = 8.58$

$a_4 = 17.5(0.7)^{4-1} = 6$

$a_5 = 17.5(0.7)^{5-1} = 4.2$

Find the total distance the ball has travelled after it hits the ground the 5th time.

$T = 17.5 + 12.25 + 8.58 + 6 + 4.2 = 48.53$

The total distance the ball has travelled after it hits the ground the 5th time is of 48.53m.

8 0

3 years ago

Just need help on 29

luda_lava [24]

Answer:

b.is the answer to your question

6 0

1 year ago

Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2

Basile [38]

The Laplace transform of the given initial-value problem

$y' 5y = e^{4t}, y(0) = 2$ is mathematically given as

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is mathematically given as

$&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}$

$\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}$

$put $s=-s \Rightarrow a_{1}=\frac{17}{9}$$

$\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}$

In conclusion, Taking inverse Laplace tranoform

$L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\$

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$