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# QUICK !!

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Please help Me on  these problems. Sorry to everybody

Aug 23, 2017
edited by BOSEOK  Aug 23, 2017
edited by BOSEOK  Oct 5, 2017

### 5+0 Answers

#1
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There are no questions here!

Aug 23, 2017
#2
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He will probably put them up after a few hours/minutes. He has done that 2 or 3 times already... Or maybe it's a lag...

Gh0sty15  Aug 23, 2017
#3
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yea, thanks Ghosty :)

Melody  Aug 23, 2017
#4
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2b^2 + b  = 6

step 1....  divide both sides by 2

b^2  + (1/2)b   = 3

step 2    .....take (1/2)  of (1/2)  = 1/4....square this = 1/16  add to both sides

b^2 + 1/2b + 1/16  = 3 + 1/16

step 3  ......factor the left side.....simplify the right side

(b + 1/4)^2  = 49/16

step 4  ......take positive/negative roots of both sides

b + 1/4  = ±√[ 49/16]

b + 1/4  =   ±7/4

step 5......subtract  1/4 from both sides

b = 7/4 - 1/4  = 6/4  = 3/2        or    b =  -7/4 - 1/4  = -8/4   = -2

The mistake was in step 2.....1/4 was added to both sides, but it should have been (1/4)^2 = 1/16 !!   Aug 23, 2017
#5
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Here's 22

f(x)  = -0.1x^2 + 3.2x -3.5

Let's  find the roots....thus

-0.1x^2 + 3.2x -3.5  = 0   multiply both sides by  -10

x^2 - 32x  + 35  = 0    subtract  35 from both sides

x^2  -  32x   =  -35       take 1/2 of 32  = 16....square this  = 256  ....add to both sides

x^2 - 32x + 256  = -35  +  256     factor and simplify

(x - 16)^2  = 221    take positive/negative roots of both sides

x - 16 =  ±√221      add 16 to both sides

x =   ±√221  + 16

So....the  two  roots are   -√  221  + 16  ≈  1.13   and √  221  + 16 ≈  30.87

And the positive difference between these two is the width of the tunnel at the bottom =

30.87 - 1.13  ≈  29.74 ft....so....the tunnel is wide enough at the bottom

The function is an upside-down parabola....the vertex occurs at the max height

To find the height...let's find the x coordinate of the vertex.....this is given by

-b  / (2a)     where  b = 3.2   and a = -.1  ....so we have

-3.2 / [ 2(-.1) ] = -3.2 / -.2  = 32 / 2  =   16

So....putting this into the original function will give us the height....and we have

-.1(16)^2  + 3.2(16)  - 3.5  =

-25.6 + 51.2 - 3.5   = 22.1  ft

However.......we need to find the width at 20 feet

So....set  the function = 20 and solve

20 =  -0.1x^2 + 3.2x -3.5  subtract 20 from both sides

Using the quadratic formula, we have that

x = ( -3.2 ±√ [ 3.2^2 - 4(-.1)(-23.5) ] ) / [ 2(-.1)] =

x = (-3.2 ±√ [  .84] ) / [ -.2 ]  =

x ≈ 11.41     and x  ≈ 20.58

So...the width at 20 feet  equals the distance between these ≈  9.17 feet

So....the tunnel is not wide enough at this point.....!!!!   Aug 23, 2017
edited by CPhill  Aug 23, 2017