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3 years ago

Can someone sub to my youttube channel pls, its called Yakobu, it would really help out :)

Mathematics

1 answer:

mestny [16]3 years ago

6 0

Answer:

yea i will what do u do ther???

Step-by-step explanation:

ohhhh gamer

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A college bookstore marks up the price that it pays the publisher for a book by 35 %. If the selling price of a book is $ 87.00

Dimas [21]

Answer:the amount that the bookstore pay the publisher for the book is $64.4

Step-by-step explanation:

Let x represent the amount that the bookstore pay the publisher for the book.

The college bookstore marks up the price that it pays the publisher for a book by 35%. This means that the value of the mark up would be

35/100 × x = 0.35 × x = 0.35x

Therefore, the amount that the bookstore is selling the book would be

x + 0.35x = 1.35x

If the selling price of a book is $ 87.00, then it means that

1.35x = 87

x = 87/1.35 = 64.4

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3 years ago

Yea I don't know this app But I heard people leaked my phone number and calling and someone saying I'm a actor yes... Please sto

pav-90 [236]

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3 years ago

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Hallar el valor de x en 3(x-2)+4=x-5

kolezko [41]

Answer:

exact form 3/2 decimal form 1.5 mixed number form 1 1/2

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2 years ago

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What is 52% of 230?

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A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

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3 years ago