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slega [8]
3 years ago
14

Please help!!

Mathematics
2 answers:
S_A_V [24]3 years ago
5 0

Answer:

50

Step-by-step explanation:

Ne4ueva [31]3 years ago
3 0

Answer:

9. 2x^{2} +5x+2

21. y=6x

23. -8

Step-by-step explanation:

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Evaluate the expression. 202 + 6(9 + 3)
andrew11 [14]

Answer:

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
HURRY!! 20 POINTS!! What is the solution set of the equation x+3/x+2=3+1/x? Note: x≠0, −2
scoundrel [369]

Answer:

i belive the answers -1

Step-by-step explanation

8 0
3 years ago
Slope-rate of change- “if there were 12,000 in the Atlantic Ocean in 1950 and there are 3,600 dolphins today, what is the yearly
jasenka [17]

\bf \begin{array}{|cc|ll} \cline{1-2} \stackrel{x}{year}&\stackrel{y}{dolphins}\\ \cline{1-2} &\\ 1950&12000\\ 2018&3600 \\&\\ \cline{1-2} \end{array}~\hspace{5em}(\stackrel{x_1}{1950}~,~\stackrel{y_1}{12000})\qquad (\stackrel{x_2}{2018}~,~\stackrel{y_2}{3600}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{3600-12000}{2018-1950}\implies \cfrac{-8400}{68}~~\approx~~ \stackrel{\textit{rate of change}}{-123.53}


-8400/68 simplifies to to -2100/17,  dolphins/year, so 2100 less dolphins every 17 years.

and we can just divide those to get the decimal amount of -123.53, which means, 123.53  less dolphins every year.

of course, there's no such a thing as 0.53 of a dolphin, so, we can say "about" 123 less dolphins per year.

5 0
3 years ago
Evaluate the definite integral. <br> 1 x4(1 + 2x5)5 dx.
Nata [24]

Answer:

Step-by-step explanation:

Given the definite integral \int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx, we to evaluate it. Using integration by substitution method.

Let u = 1-2x⁵ ...1

du/dx = -10x⁴

dx = du/-10x⁴.... 2

Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;

= \int\limits {\dfrac{x^4}{u^5} } \, \dfrac{du}{-10x^4}

= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} }  \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du }  \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}]  \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\

= \dfrac{u^{-4}}{40} \\\\\\= \dfrac{1}{40u^4} +C

substitute u = 1-2x⁵ into the result

= \dfrac{1}{40(1-2x^5)^4} +C

Hence\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx = \dfrac{1}{40(1-2x^5)^4} +C

5 0
3 years ago
If a fire is burning a building and it takes 2 hours to put it out using 2000 gal./min., what minimum diameter cylindrical tank
strojnjashka [21]

Answer:

The minimum diameter of the cylindrical tank needed to store the quantity needed to put out the fire is approximately 58.415 feet.

Step-by-step explanation:

A gallon equals 0.134 cubic feet. First, we determine the amount of water (Q), measured in cubic feet, needed to put out the fire under the assumption that water is consumed at constant rate:

Q = \dot Q \cdot \Delta t (1)

Where:

\dot Q - Volume rate, measured in feet per minute.

\Delta t - Time, measured in minutes.

If we know that \dot Q = 2000\,\frac{gal}{min} and \Delta t = 120\,min, then the amount of water is:

Q = \left(2000\,\frac{gal}{min} \right)\cdot (120\,min) \cdot \left(0.134\,\frac{ft^{3}}{gal} \right)

Q = 32160\,ft^{3}

And the diameter of the cylindrical tank based on the capacity found above is determined by volume formula for a cylinder:

Q = \frac{\pi}{4}\cdot D^{2}\cdot h (2)

Where:

D - Diameter, measured in feet.

h - Height, measured in feet.

If we know that Q = 32160\,ft^{3} and h = 12\,ft, then the minimum diameter is:

D^{2} = \frac{4\cdot Q}{\pi\cdot h}

D = 2\cdot \sqrt{\frac{Q}{\pi\cdot h} }

D = 2\cdot \sqrt{\frac{32160\,ft^{3}}{\pi\cdot (12\,ft)} }

D \approx 58.415\,ft

The minimum diameter of the cylindrical tank needed to store the quantity needed to put out the fire is approximately 58.415 feet.

8 0
3 years ago
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