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For how long is the cannonball above a height of 5 meters?

Hello Guest!

\( h(t) = -4.9t^2 + 14t - 0.4\)

\(-4.9t^2 + 14t - 0.4=5\\ -4.9t^2 + 14t - 5.4=0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(t= {-14 \pm \sqrt{196-4\cdot 4.9\cdot 0.4} \over 2\cdot (-4.9)}\\ t=\dfrac{-14\pm 13.7171}{-9.8}\)

\(t\in \{0.02886s,2.8283s\}\)

2.8283s - 0.0288s = 2.799s long is the cannonball above a height of 5 meters.

  !

First I worked out what the inverse of 3 is mod 11

Let the inverse of 3 be x

3x=1 (mod11)

3x=11k+1    

3x=9k+(2k+1)

x=3k+(2k+1)/3

Let b=(2k+1)/3

3b=2k+1

2k=2b+(b-1)

k=b+(b-1)/2

Let c=(b-1)/2

2c=b-1

b=2c+1

so   k=(2c+1)+c = 3c+1

and

x= 3(3c+1)+b = 9c+3+2c+1 = 11c+4 = 4 (mod11)

So the inverse of 3 mod 11 is 4 mod11

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Now I went through the same procedure and got that the inverse of 5 mod11 was -2 mod11  

( I have not checked my working)

(4+-2)mod 11 = 2 mod 11

NowI went through the whole procedure again and found that the inverse of 2 mod 11 is  6 mod 11

So I got a is equivalent to 6 (mod11)

I have not checked my working and maybe there is an easier way to do it.

Hint: First, take the derivative with respect to time of both sides of the equation. Then, you can use substitution to solve. 

In trapezoid EFGH, EF is parallel to GH and P is the midpoint of line segment EH. If the area of triangle PEF  is 18, and the area of triangle PGH  is 30, then find the area of trapezoid EFGH.

[EFGH] = 96

Quadrilateral ABCD  is a parallelogram. Let E  be a point on AB, and let F be the intersection of lines DE and BC. The area of triangle EBF is 1 and the area of triangle EAD  is 20. Find the area of parallelogram ABCD.

Let quadrilateral ABCD be a rectangle.

AE = 10

AD = 4

BE = √5

BF = √0.8

[ABCD] = 4(10 + √5)

yeah

pls ellaborate 

The answer is 12 seconds.

3x^2 - 4x + 12.

Let AB and CD be chords of a circle, that meet at point Q inside the circle. If AQ=6, BQ=12, and CD=38, then find the minimum length of CQ.

AQ * BQ = 72

CQ(38 - CQ) = 72

CQ = 2

Angle A = 45º       

AD = DE = 3√2

CD = 12 / 3√2 = 2√2

[BCE] = 6 + (2√2)² / 2

[ABC] = [ADE] + [CDE] + [BCE] 

MAX   occurs at   t = - b/2a = - 8 /(2*-1) = 4 

    sub in t = 4 to find the max value is   12

(there is no minimum value....   - infinity)

f(x) = 5x^3 - 6x^2 + 4x + 2.

$\sin \theta = \frac{3}{5}$

You get the minimum at t = -8, which gives you a minimum value of -4.

In terms of pi, what is the area of the circle defined by the equation $2x^2+2y^2+10x-6y-18=20?

Hello Guest!

\(2x^2+2y^2+10x-6y-18=20\\ x^2+y^2+5x-3y-19=0\\ \ 27.5=19+2.25+6.25\\ (y^2-3y+2.25)=27.5-(x^2+5x+6.25)\\ (y-1.5)^2=27.5-(x+2.5)^2 \)       \(r^2=27.5\\ P_{centr}\ [-2.5,1.5]\)

\(y=\pm \sqrt{27.5-(x+2.5)^2 }+1.5\)

\(r=\sqrt{27.5}\\ A=r^2\pi \)

\(A=27.5\ \pi\)

  !    

First of we have the equation, 

6x + 4x - 6 = 24 + 9x

Grouping all the x's onto one side we get, 

6x + 4x - 9x = 24 + 6

10x -9x = 30

x = 30 

Checking our answer we get,

6(30) + 4(30) - 6 = 24 + 9(30)

180 + 120 - 6 = 24 + 270

300 - 6 = 294

294 =294        this means our answer is correct. 

Hope this was helpful 

The average of x + 6,6x + 2, and 2x is 4x-7. What is x?

In order to find the average of the numbers you have to add all of them and then divide by the amount of numbers you have to average. 

First,

[(x + 6) + (6x+2) + 2x ]/3 = 4x-7 

Getting rid of the denominator, by multiplying by 3 on both sides,

[(x + 6) + (6x+2) + 2x ] = 12x - 21

Grouping all the x's 

[(x + 6x + 2x) + 8] = 12x - 21

[9x+8] = 12x -21 

-3x = -29

x = 29/3

Hope I didin't make any silly mistakes lol. 

https://web2.0calc.com/questions/tangent-question-with-inscribed-angles-please-explain

3 min/lap

h(t) = -4.9t^2 + 14t - 0.4

7 = -4.9t^2 + 14t - 0.4

0 = - 04.9t^2 + 14t - 7.4     Use Quadratic formula to solve for two values of 't'......time between is the answer

Take log of both sides

 x log 4  = log 1200

x = log1200/log 4 = 5.1144

Two.