All Answers 359137 Answers

Here's a numerical approach:

.So x ≈ 2.162939

.

sin u = -4/5 , pi < u < 3pi/2

answers

sin2u = 24/25

cos2u = -17/25  not correct

tan 2u = 4sqrt21/ 17

how to do?

\(sin \ 2u=2sin \ u \cdot cos \ u\\cos \ u=\sqrt{1-sin^2u}\\sin\ 2u=2sin\ u\cdot\sqrt{1-sin^2u}\)

\(sin \ 2u=-\frac{8}{5}\cdot\sqrt{1-\frac{16}{25}}=-\frac{8}{5}\cdot -\sqrt{\frac{9}{25}}=-\frac{8}{5}\cdot -\frac{3}{5}\)

\( sin\ 2u=\frac{24}{25}\) 

\(cos \ 2u=1-2sin^2u=1-2\cdot\frac{16}{25}\)

\(cos\ 2u=-\frac{7}{25}\)

\(tan \ 2u=\ \frac{2\cdot tan\ u}{1-tan^2u}\\tan\ u=\pm\ \frac{sin\ u}{\sqrt{1-sin^2u}}\)

\(tan^2u=\frac{sin^2u}{1-sin^2u}=\frac{\frac{16}{25}}{1-\frac{16}{25}}=\frac{16\cdot 25}{25\cdot 9}\)

\(tan\ u =\frac{4}{3}\)

\(tan\ 2u=\frac{\frac{8}{3}}{1-\frac{16}{9}}=-\ \frac{8\cdot 9 }{3\cdot 7}\)

\(tan\ 2u=-\ \frac{24}{7}\)

  !

Solve for X given : 5.52 = -ln( -ln( -(1-1/X) ))

\(\begin{array}{|rcll|} \hline 5.52 &=& -\ln \Big( -\ln ( -(1-\frac{1}{x} ) ) \Big) \\ 5.52 &=& -\ln \Big( -\ln( \frac{1}{x}-1 ) \Big) \\ -5.52 &=& \ln \Big( -\ln( \frac{1}{x}-1 ) \Big) \quad & | \quad e^{()} \\ e^{-5.52} &=& -\ln( \frac{1}{x}-1 ) \\ -e^{-5.52} &=& \ln( \frac{1}{x}-1 ) \quad & | \quad e^{()} \\ e^{(-e^{-5.52})} &=& \frac{1}{x}-1 \\ 1+e^{(-e^{-5.52})} &=& \frac{1}{x} \\ \dfrac{1}{1+e^{(-e^{-5.52})} } &=& x \\ x &=& \dfrac{1}{1+e^{(-e^{-5.52})} } \\ x &=& \dfrac{1}{1+e^{(-0.00400584794)} } \\ x &=& \dfrac{1}{1+0.99600216476 } \\ x &=& \dfrac{1}{1.99600216476 } \\\\ \mathbf{x} & \mathbf{=} & \mathbf{0.50100146065} \\ \hline \end{array}\)

Here are the steps:

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find angle c with trig

\(\begin{array}{|rcll|} \hline \tan(B) &=& \dfrac{c \cdot \sin(A)}{b-c\cdot \cos(A)} \quad & | \quad c = 75\ cm \quad b = 90\ cm \quad A = 20^{\circ} \\\\ \tan(B) &=& \dfrac{75 \cdot \sin(20^{\circ})}{90-75\cdot \cos(20^{\circ})} \\\\ \tan(B) &=& \dfrac{75 \cdot 0.34202014333}{90-75\cdot 0.93969262079} \\\\ \tan(B) &=& 1.31390875033 \\\\ B &=& \arctan(1.31390875033) \\\\ \mathbf{B} & \mathbf{=} & \mathbf{52.7256774464^{\circ}} \\ \hline \end{array}\)

1- Convert 5 radians to degrees.  Round answer to the nearest tenth of a degree.

2- Convert 6/x^4 into fractional or root form.

3- Find the solution set of 18x - 3x^2 > 0

7,5*x+15*(230-x)*0,58+15*(230-x)*0,58*0,58+7,5*x*0,72=6525

\(7,5*x+15*(230-x)*0,58+15*(230-x)*0,58*0,58+7,5*x*0,72=6525\)

\(Man, you \ can \ do \ it \ yourself \ with \ a \ pocket \ calculator \ and \ some \ diligence.\)

  !

1)The center of a circle is at (−3, 1) and its radius is 9.

What is the equation of the circle?

(x+3)²+(y−1)²=81

3)

The equation of a circle is (x+4)²+(y−5)²=121.

What is the center and radius of the circle?

center: (−4, 5); radius: 11

4)

What is the greatest distance, in feet, a person could be from the lamp and be detected?

3 ft

5)

What is the diameter, in meters, of the circular area that gets watered by the sprinkler?

12 m

Which graph shows the graph of a circle with equation x2+(y+5)2=25?

1) ellipse

2) circle

4) parabola

5) line and degenerate parabola

* I thought it was degenerate hyperbola because of the lines that seem to make sense to me, but it was degenerate parabola

Reference for help with Conic Sections:

https://quizlet.com/22927630/conic-sections-flash-cards/

4 - 8 sin^2 (x)        factor

4  [ 1  - 2 sin^2 (x)   ]      remember that  1 - 2sin^2x  =  cos (2x)

So  we end up with

4  cos (2x)

We can use similar triangles to find the radius, R, when the height of the sand  = 30mm

So we have...

10 /24  = R / 30      multiply both sides by 30

30 * 10  / 24   =  12.5 mm  

So the volume of the sand  is now  = [ (1/3) pi ( 12.5)^2 * 30 ]  m^3

And since the sand flows out at 50mm^3 / sec......the time you have to answer the question =

[ (1/3) pi ( 12.5)^2 * 30 ] / 50   ≈  98 sec 

learning math because it helps with problem solving skills. even if you don't become an engineer or a math teacher, math forces you to look at problems at different angles. 

There are several approaches, but here's mine

Let's find  AD  using the Law of Sines

AD / sin ABD  = AB / sin BDA

AD /  sin(70)  =  75 / sin (90)

AD  =  75 sin (70)  

So....DC  =   90  - 75sin (70)  

Similarly.... we can find BC  as  75 sin (20)   

So......using the tangent

Tan ( C )   =    BC / DC

Tan (C)  =   [ (75) sin (20) /  (   90  - 75sin (70) ) ]

Using the arctangent

arctan   [ 75 sin (20) /  (   90  - 75sin (70) ) ]  = C  ≈  .92 rads  ≈  52.7°

Is this what you needed? Or did I type it differently than what you meant?

Thanks Chris, I didn't see the interger bit.  :/

So yes you need three integers that are as close as posible to the same value, that multiply to give 100.

Yeah, pretty much just to make you smarter, but so you can apply it to real-life situations in the future when you are doing whatever. It would be pretty helpful to know how to measure the dimensions of parts for a house so it doesn't fall over, or to advance the field of science with complex calculations.

I assume you  want to write the equation for this line......we have

y - (-2)  =  (3/5) ( x - 0)

y + 2  =  (3/5)x         subtract 2 from both sides

y  =  (3/5)x  -  2

Got no idea how to step through this  [ maybe someone else does ???  ]

WolframAlpha  gives the answer  as    X ≈ 0.501001460646335

https://www.wolframalpha.com/input/?i=5.52+%3D+-ln(-ln+-+(1-(1%2FX)))

yep, i realised after your comment that the question was a graohic calculator question! 

thank you so much for your help anyways, gave me a real stump. 

Using Heron's formula  to find the area of one of the triangle bases ....

The semi-perimeter, s, =     [ 19 + 19 + 26.9]  / 2   = 32.45

And the area of one base  =   sqrt  [ ( 32.45) (32.45 - 19)^2 ( 32.45 - 26.9) ]  ≈ 180.5  in^2

So.....the surface  area  of one prism =

Area of the bases + Area of the sides  =

2 (180.5) + 41 [ 19 + 19 + 26.9]   = 3021.9 in ^2

And doubling this gives the total surface area of both prisms combined  ≈ 6043.8 in^2

If we glued the  26.9 in  sides together  we would have a rectangular prism with square bases of 19 inches on each side

So.......the surface area of such a prism  =

Area of both bases + Area of the sides

2(19)^2  +  41 ( 19 * 4 )   = 3838 in^2 

I'm just  playing around with this......I don't know if it's correct....

L * W * H

Volume

Surface Area

1 * 1 * 100  m^3

2 ( 1  + 100 + 100) =  400  m^2

2 * 2 * 25 =  100m^3

2 ( 4 + 50 + 50)  = 208  m^2

2 * 5 * 10   = 100 m^3

2( 20 + 10 + 50)  =  160 m^2

5 * 5 * 4 =  100 m^3

2(25 + 20 + 20)  =  130 m^2

WolframAlpha  shows that the surface area is minimized when each dimension ≈ 4.64159 m

And the surface area ≈ 129.266 m^2 

Which is exactly what Melody got........

But.....since each dimension must be an integer.........

I'm assuming that my last answer must be pretty close, since the surface area  = 130 m^2

That is.....we have  a square base of 5m on each side  and a height of 4 m

what would the lowest surface area for a prism with a volume of 100 meters cubed??? 

(using only whole numbers for dementions)

The lowest surface area would be if the sides were all the same.

side length = 100^(1/3) = 4.641588833612778m        

So the surface area would be   6*(100^(1/3))^2 = 129.2660814019129736   cm^2

this is the same as 6*4.641588833612778^2

or

the same as      6*(100^(2/3))

My friend Google secretly told me that the volume of a sphere can be found using this formula: \(V = \frac{4}{3}\pi r^3\)

So, to find the volume, plug in the radius of the sphere (or half the diameter, if given) and just solve it from there.