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Gnesinka [82]
3 years ago
11

Can some one please help me find the answer?

Mathematics
2 answers:
alexandr1967 [171]3 years ago
8 0

Answer:

V = 20.64573mm^3

Step-by-step explanation:

a = 15/5 = 3

h = 4

V = (5/12) tan(54) h a^2

V = (5/12) tan(54) (4) (3)^2

V = 20.64573

sergiy2304 [10]3 years ago
6 0
516.14mm I think that’s the answer
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What's the median of these numbers:<br> 14,20,24,16,20,18
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The answer to this is 19. The reason why is because the median falls between 18 and 20. After that, you have to take the middle number that falls between 18 and 20. In this case, 19. 
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A function f(x)=3x+12.
seraphim [82]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2822258

_______________


•  Function:   f(x) = 3x + 12.


A.  Finding the inverse of f.

The composition of f with its inverse results in the identity function:

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B.  Verifying that the composition of f and g gives us the identity function:

•  \mathsf{(f\circ g)(x)}

\mathsf{=f\big[g(x)\big]}\\\\\\ \mathsf{=3\cdot \left(\dfrac{x}{3}-4\right)+12}\\\\\\&#10;\mathsf{=\diagup\hspace{-7}3\cdot \dfrac{x}{\diagup\hspace{-7}3}-3\cdot 4+12}\\\\\\&#10;\mathsf{=x-12+12}\\\\&#10;\mathsf{=x\qquad\quad\checkmark}


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\mathsf{=g\big[f(x)\big]}\\\\\\ \mathsf{=\dfrac{f(x)}{3}-4}\\\\\\ \mathsf{=\dfrac{3x+12}{3}-4}\\\\\\&#10;\mathsf{=\dfrac{\diagup\hspace{-7}3\cdot (x+4)}{\diagup\hspace{-7}3}-4}\\\\\\&#10;\mathsf{=x+4-4}\\\\&#10;\mathsf{=x\qquad\quad\checkmark}

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C.  Since f and g are inverse, then

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•  Call h the compositon of f and g. So,

h(x) = (f o g)(x)

h(x) = x


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I hope this helps. =)

5 0
3 years ago
3) 10 - 1 * 10 / 5 Can Someone Explain How To Solve This Problem For A Example​
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3 years ago
A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of
belka [17]

Let <em>s(t)</em> be the amount of salt in the tank at time <em>t</em>. Then <em>s(0)</em> = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (<em>s(t)</em>/300 lb/gal) = <em>s(t)</em>/150 lb/min

so that the net rate at which salt is exchanged through the tank is

d<em>s(t)</em>/d<em>t</em> = 12 - <em>s(t)</em>/150 … (lb/min)

Solve for <em>s(t)</em>. The DE is separable, so we have

d<em>s</em>/d<em>t</em> = 12 - <em>s</em>/150

150 d<em>s</em>/d<em>t</em> = 1800 - <em>s</em>

150/(1800 - <em>s</em>) d<em>s</em> = d<em>t</em>

Integrate both sides to get

-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>

Solve for <em>s</em> :

ln|1800 - <em>s</em>| = -<em>t</em>/150 + <em>C</em>

1800 - <em>s</em> = exp(-<em>t</em>/150 + <em>C </em>)

1800 - <em>s</em> = <em>C</em> exp(-<em>t</em>/150)

<em>s</em> = 1800 - <em>C</em> exp(-<em>t</em>/150)

Now given that <em>s(0)</em> = 50, we solve for <em>C</em> :

50 = 1800 - <em>C</em> exp(-0/150)

50 = 1800 - <em>C</em>

<em>C</em> = 1750

Then the amount of salt in the tank at any time <em>t</em> ≥ 0 is

<em>s(t)</em> = 1800 - 1750 exp(-<em>t</em>/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for <em>t</em> in

100 = 1800 - 1750 exp(-<em>t</em>/150)

1700 = 1750 exp(-<em>t</em>/150)

34/35 = exp(-<em>t</em>/150)

ln(34/35) = -<em>t</em>/150

<em>t</em> = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for <em>s(t)</em>, we could have instead stopped with

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Solve for <em>C</em> - this <em>C</em> <u>is not</u> the same as the one we found using the other method. <em>s(0)</em> = 50, so

-150 ln|1800 - 50| = 0 + <em>C</em>

<em>C</em> = -150 ln|1750|

==>   <em>t</em> = 150 ln(1750) - 150 ln|1800 - <em>s</em>|

Then <em>s(t)</em> = 100 lb when

<em>t</em> = 150 ln(1750) - 150 ln(1700)

<em>t</em> = 150 ln(1750/1700)

<em>t</em> = 150 ln(35/34)

<em>t</em> ≈ 4.348

5 0
3 years ago
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