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3 years ago

9

What is the solution of

rmula1" title="\frac{x+2}{4} = \frac{x+8}{3}" alt="\frac{x+2}{4} = \frac{x+8}{3}" align="absmiddle" class="latex-formula">

1 answer:

Lina20 [59]3 years ago

6 0

Answer:

x = -26

Step-by-step explanation:

Multiply both sides by the lowest common multiple of 4 and 3, which is 12.

3(x+2) = 4(x+8)

3x + 6 = 4x + 32 --- distribute the 4 and 3

6 = x + 32 --- subtract 3x from both sides

-26 = x --- subtract 32 from both sides

x = -26

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What is the equation for the line

Darya [45]

That depends the line
1, If the line is not parallel to Oy, so its equation is y=ax+b
2, Or its equation is x=m

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4 years ago

If points c, d, and e are on a line and cd =20 and ce = 32 what are the possible values of de

vivado [14]

In this question it is given that the values of CD and CE are 20 and 32 respectively.

Since the length of CE is greater then of CD, so C and D could not be the end points.

Therefore, either D and E are the endpoints or C and E are the endpoints.

If D and E are the endpoints, then DE is the sum of 20 and 32 which is 52.

And if C and E are the endpoints, then DE is the difference of CE and CD that is 12.

So the possible values of DE are 12 and 52.

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3 years ago

A motorboat that travels with a speed of 20 km/hour in still water has traveled 20 km against the current and 180 km with the cu

Gala2k [10]

Recall your d = rt. distance = rate * time.

let's say the speed of the current is "c".

keeping in mind that, when the boat is going against the current, the boat is not really going at 2kmh, but instead at " 20 - c ", because the current is subtracting speed from it.

Likewise when is going with the current, is going at a speed of " 20 + c ".

if the boat took say "t" hours on the way over, on the way back it took the slack from the whole 8 hours and "t", namely it took " 8 - t " hours.

$\bf \begin{array}{lccclll}
&\stackrel{km}{distance}&\stackrel{kmh}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{against the current}&20&20-c&t\\
\textit{with the current}&180&20+c&8-t
\end{array}
\\\\\\
\begin{cases}
20=(20-c)(t)\implies \frac{20}{20-c}=t\\
180=(20+c)(8-\stackrel{\downarrow }{t})\\
--------------\\
180=(20+c)\left( 8- \stackrel{\downarrow }{\frac{20}{20-c}}\right)
\end{cases}
\\\\\\
180=(20+c)\left( \cfrac{(160-8c)-20}{20-c} \right)$

$\bf 180=(20+c)\left( \cfrac{140-8c}{20-c} \right)\implies 180(20-c)=(20+c)(140-8c)
\\\\\\
3600c-180c=2800-160c+140c-8c^2
\\\\\\
3600c-180c=2800-20c-8c^2
\\\\\\
3600c-180c-2800+20c+8c^2=0\implies 8c^2-160c+800=0
\\\\\\
c^2-20c+100=0\implies (c-10)(c-10)=0\implies \boxed{c=10}$

8 0

3 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

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3 years ago

At the frozen yogurt shop, a machine fills cups with 4 ounces of frozen yogurt before adding the toppings. After the cups are fi

Otrada [13]

What are the options?

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3 years ago