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MAVERICK [17]
3 years ago
10

You use 4 gallons of water on 14 plants in your garden.

Mathematics
2 answers:
GrogVix [38]3 years ago
4 0

Answer:

35 plants

Step-by-step explanation:

For this case we can make the following rule of three:

4 gallons --------------> 14 plants

x --------------------------> 35 plants

From here, we clear the value of x.

We have then:

therefore, 10 gallons are needed to irrigate 35 plants.

Answer:

it will take 10 gallons to water 35 plants

Reptile [31]3 years ago
4 0
Answer: 10 gallons

4 gallon/14 plants= x gallons/ 35 plants

35*4=140/14=x

x=10 gallons
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The perimeter of a rectangle is 24 meters. If the length of the rectangle is 8 meters, what is the width (in meters)?
Serhud [2]

Answer:

8 meters

Step-by-step explanation:

8 + 8 + 8 + 8 = 24 which is the perimeter, just trust me

7 0
3 years ago
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Find the midpoint of AB if A (-3 , 8) and B (-7 , -6).
matrenka [14]

Answer:

(-5,1)

Step-by-step explanation:

the formula is

C=((-3-7)/2,(8-6)/2)

therefore,

C=(-5,1)

HOPE IT WILL HELP!!

8 0
3 years ago
Hii:) anyone able to help with practice 6? Thanks:)!
Leno4ka [110]

Answer:

The answer is

<h2>x =  \frac{ {e}^{ \frac{3}{ ln(2) } } }{4}</h2>

Step-by-step explanation:

<h3>ln(2)  \times  ln(4x)  = 3</h3>

<u>Divide both sides of the equation by ln 2</u>

That's

<h3>\frac{ ln(2) ln(4x)  }{  ln(2)  }  =  \frac{3}{ ln(2) }</h3>

We have

<h3>ln(4x)  =  \frac{3}{ ln(2) }</h3>

Covert the logarithm into exponential form using the property

<h3>ln(x)  = b  \:  \:  is \: \: \:  the \: \: \: same \: \: \: as  \:  \:  \: \:x =  {e}^{b}</h3>

Are all the same

So

<h3>ln(4x)  =  \frac{3}{ ln(2) } </h3>

is the same as

<h3>4x =  {e}^{ \frac{3}{ ln(2) } }</h3>

Divide both sides by 4

We have the final answer as

<h3>x =  \frac{ {e}^{ \frac{3}{ ln(2) } } }{4}</h3>

Hope this helps you

3 0
3 years ago
The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section
Ber [7]
Given that y_1=e^{2x/3}, we can use reduction of order to find a solution y_2=v(x)y_1=ve^{2x/3}.

\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}
\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}

\implies9y''-12y'+4y=0
\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0
\implies9v''-3v'=0

Let u=v', so that

9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0
e^{-x/3}u'-\dfrac13e^{-x/3}u=0
\left(e^{-x/3}u\right)'=0
e^{-x/3}u=C_1
u=C_1e^{x/3}

\implies v'=C_1e^{x/3}
\implies v=3C_1e^{x/3}+C_2

\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}
\implies y_2=3C_1e^x+C_2e^{2x/3}

Since y_1 already accounts for the e^{2x/3} term, we end up with

y_2=e^x

as the remaining fundamental solution to the ODE.
7 0
3 years ago
Write a cosine function that has an amplitude of 5, a midline of 2 and a period of \frac{2\pi}{5} 5 2π ​ .
vlabodo [156]

Answer:

f(x) = 5*cos(5*x) + 2

Step-by-step explanation:

A general cosine function is written as:

f(x) = A*cos(ω*x) + M

where:

A is the amplitude

M is the midline

ω is the angular frequency.

In this case we know that:

The amplitude is 5, then A = 5

The midline is 2, then M = 2

The period is 2*pi/5

The relation between the angular frequency ω and the period T is:

ω = 2*pi/T

Then if the period is T = 2*pi/5, replacing that in the above equation we find that:

ω = 2*pi/T = 2*pi/(2*pi/5) = 5

ω = 5

Then the function is:

f(x) = 5*cos(5*x) + 2

7 0
3 years ago
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