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3 years ago

5

Mario earns money mowing his neighbors' lawns. The revenue for mowing x lawns is r(x) = 25x. Mario's cost for gas and the mower

rental is c(x) = 6x + 15. his profit from mowing is x lawns is p(x) = (r-c)(x). What is p(x)

2 answers:

Arturiano [62]3 years ago

5 0

I know for this one the answer is p(x)= 19x-15

m_a_m_a [10]3 years ago

4 0

Answer:

The value of p(x) is 19x - 15.

Step-by-step explanation:

Given,

The revenue mowing x lawns is, r(x) = 25x.

Also, Mario's cost for gas and the mower rental is c(x) = 6x + 15.

Hence, his profit from mowing is x lawns is,

p(x) = (r-c)(x) = r(x) - c(x) ( Subtracting functions property )

By substituting values,

p(x) = 25x - ( 6x+ 15 )

= 25x - 6x - 15 ( Associative property )

= 19x - 15

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&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~{{ a}} &,&{{ c}}~) 
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\end{array}
\\\\\\
% slope = m
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3 years ago

Suppose a certain population satisfies the logistic equation given by dP

Ksenya-84 [330]

Answer:

The population when t = 3 is 10.

Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

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Using variable separable method we get

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Using partial fraction

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Using these values the equation (1) can be written as

$\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt$

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On simplification we get

$\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C$

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$\frac{1}{10}(\ln \frac{1}{10-1})=0+C$

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$\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})$

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$\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}$

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Reciprocal it

$\dfrac{10-P}{P}=9e^{-10t}$

$P(t)=\dfrac{10}{1+9e^{-10t}}$

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3 years ago

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denis23 [38]

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3 years ago

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vovangra [49]

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3 years ago

Read 2 more answers

In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some poin

FrozenT [24]

Answer: The length of segments between this point and the vertices of greater base are $7\frac{5}{7}$ and 18.

Step-by-step explanation:

Let ABCD is the trapezoid, ( shown in below diagram)

In which AB is the greater base and AB = 18 DC= 11, AD= 3 and BC = 7

Let P is the point where The extended legs meet,

So, according to the question, we have to find out : AP and BP

In Δ APB and Δ DPC,

∠ DPC ≅ ∠APB ( reflexive)

∠ PDC ≅ ∠ PAB ( By alternative interior angle theorem)

And, ∠ PCD ≅ ∠ PBA ( By alternative interior angle theorem)

Therefore, By AAA similarity postulate,

$\triangle APB\sim \triangle D PC$

Let, DP =x

⇒ $\frac{3+x}{18} = \frac{x}{11}$

⇒ 33 +11x = 18x

⇒ x = 33/7= $4\frac{5}{7}$

Thus, PD= $4\frac{5}{7}$

But, AP= PD + DA

AP= $4\frac{5}{7}+3 =7\frac{5}{7}$

Now, let PC =y,

⇒ $\frac{7+y}{18} = \frac{y}{11}$

⇒ 77 + 11y = 18y

⇒ y = 77/7 = 11

Thus, PC= 11

But, PB= PC + CB

PB= 11+7 = 18

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3 years ago