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3 years ago

What is the slope of a line parallel to a line with equation 2x - 5y = 9

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

5 0

First take and get the slope of the line. If the slope is equal to an option then that's your answer. Hope I helped!

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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

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

3 0

3 years ago

how much time does it take for a bird flying at a speed of 45 km per hour to travel distance of 1800 km

Andrew [12]

Answer:

40 hours

Step-by-step explanation:

1800 divided by 45 equals 40 hours. To check your answer you can do 45 times 40.

Hope this helped.

3 0

3 years ago

Read 2 more answers

Solve 2/3 (5x4) / 3 + 6/7=

Anon25 [30]

3 37/81, i think is it

5 0

3 years ago

Read 2 more answers

Simplify and answer the boxes.

s2008m [1.1K]

Answer:

$\huge\boxed{\dfrac{x^2+9^2}{x-3y}+\dfrac{6xy}{3y-x}=x-3y}$

Step-by-step explanation:

Domain:

$x-3y\neq0\Rightarrow x\neq3y$

$\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{3y-x}=\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{-(x-3y)}\\\\=\dfrac{x^2+9y^2}{x-3y}-\dfrac{6xy}{x-3y}=\dfrac{x^2+9y^2-6xy}{x-3y}\\\\=\dfrac{x^2-2(x)(3y)+(3y)^2}{3y-x}=\dfrac{(x-3y)^2}{3y-x}\\\\=\dfrac{\bigg[-1(3y-x)\bigg]^2}{3y-x}=\dfrac{(-1)^2(3y-x)^2}{3y-x}\\\\=\dfrac{1(x-3y)(x-3y)}{x-3y}=x-3y$

Used:

The distributive property: a(b + c) = ab + ac

(a - b)² = a² - 2ab + b²

6 0

2 years ago

Help me solve for X <br> Please

frutty [35]

Step-by-step explanation:

18+3=x+3

21=x+3

x=21-3

x=18

<h2>stay safe healthy and happy...</h2>

4 0

3 years ago