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Advocard [28]
2 years ago
14

*PLEASE HURRY WILL GIVE BRAINLY* A company's profits (in dollars) are modeled by the function p(x) = —x^2+ 1100x - 300000, where

x is the number of units produced. To determine the minimum number
of units that must be produced in order to earn profits of $590, the company plots the parabola of y = —x^2 1100x -- 300000 and the line x = 590, finds the intersection

of the two, and concludes that 900 units must be produced. Is the company correct? Explain, rounding values to the nearest whole number if necessary.
Mathematics
1 answer:
LekaFEV [45]2 years ago
4 0
No, it’s not correct. The y-axis on the graph represents the profits p(x) so the minimum number of units produced should be when the a horizontal line at y = 590 first intersects the parabola drawn left to right, and not a vertical line at x = 590 because that represents the profit as 590 units are produced.
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Steps
1. Do the point slope form
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klasskru [66]

Answer:

110 degrees, 84 degrees

Step-by-step explanation:

Picture 1:

See attached image for the angles I am referencing.

measure of angle a= measure of the 110 degree angle because of corresponding angles

measure of angle b=measure of angle a because of vertical angles

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3 years ago
D/d{cosec^-1(1+x²/2x)} is equal to​
SIZIF [17.4K]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

Let assume that

\rm :\longmapsto\:y =  {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

We know,

\boxed{\tt{  {cosec}^{ - 1}x =  {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}

So, using this, we get

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 +  {x}^{2} } \bigg)

Now, we use Method of Substitution, So we substitute

\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z =  {tan}^{ - 1}x}

So, above expression can be rewritten as

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 +  {tan}^{2} z} \bigg)

\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)

\rm\implies \:y = 2z

\bf\implies \:y = 2 {tan}^{ - 1}x

So,

\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x

Thus,

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

\rm \:  =  \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)

\rm \:  =  \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)

\rm \:  =  \: 2 \times \dfrac{1}{1 +  {x}^{2} }

\rm \:  =  \: \dfrac{2}{1 +  {x}^{2} }

<u>Hence, </u>

\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) =  \frac{2}{1 +  {x}^{2} }}}}

<u>Hence, Option (d) is </u><u>correct.</u>

6 0
2 years ago
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Answer:

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Step-by-step explanation:

8 0
2 years ago
HELP PLZ DUE TM!!!! 20 POINTS!!!
Flauer [41]

 

\displaystyle\bf\\m \overset{\frown}{HE}=360-m \overset{\frown}{HL}-m \overset{\frown}{EV}-m \overset{\frown}{VL}\\m\overset{\frown}{HE}=360^o-40^o-130^o-110^o=360^o-280^o=80^o\\\\m\widehat{EYH}=m\widehat{EYV}=\frac{m \overset{\frown}{EV}-m\overset{\frown}{HE}}{2}=\frac{130^o-80}{2}=\frac{50^o}{2}=\boxed{\bf25^o}

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2 years ago
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