All Answers 356003 Answers

The relationship EM^2=DM*MF is very helpful...

DF=\(4\sqrt{2}\), so \(DM=2\sqrt{2}\) and \(MF=2\sqrt{2}\).

\(EM^2=2\sqrt2*2\sqrt{2}\)

\(EM^2=8\)

Area  of square  =   500^2

Area  of circle   =  pi (250)^2

Area not watered  =   500^2  -  pi (250)^2  

Can you  take it from here, helpppp???

[ I'll check your answer if you want me to  ]

Let Antonio's age be A and Sarah's be S.  \(2A + 3S = 34 , S=5A\) .  Substituting 5A for S in the first equation we get 17A = 34 and A =2 . Then Sarah's age is 5 times that so it is 10. 

Recognize that MBCN is a trapezoid-and the area can be found by: 1/2 * (sum of the bases) * (height).

The height is \(\frac{40}{8}=5\) units,

Thus, the area of MBCN)(the trapezoid) is \(\frac{1}{2}*6*5=15\) units squared.

Thanks, Cal   !!!!

As Cal found,  the  other two test scores  sum to 174

If we divide  174  by 2  we get  87

But we can't  have  the same scores

And  the two scores  can't be  88  and 86  because we already have a score of 88

But note....if we let one of the scores  =  89, then the other could  be 174 - 89  =  85

So  we  have

70  73  85  88 89  !!!

Both triangles are 30-60-90 triangles. Using triangle properties, we get that AD is 3 and DB is 1. We also know that CB is 2.  AC is \(2\sqrt{3}\). The area is 0.5*base*height so it is \(2\sqrt{3}\)

By the definition of 30 - 60 - 90 triangles...AC=\(2\sqrt{3}\) and \(AD=\sqrt{3}\).

Triangle CDB is also a 30-60-90 degree triangle...angle B is sixty(60) degrees.

That means BD=1, and CB=2

..Thus, the area of the triangle is \(\frac{1}{2}*2*2\sqrt{3}=2\sqrt{3}\)

Excellent, MathisCool  !!!!

First, we solve f^-1, which can be done by setting f(y)=x...

Thus, we have 3y^2=x

\(y=\frac{\sqrt{3x}}{3}\).

Thus, f^-1(12) is \(y=\frac{\sqrt{3*12}}{3}=\pm2.\)

To find the inverse, switch x and y in the equation, making it \(x=3y^2\) and plugging in 12, you get 2 or -2. 

This article will help :D

Thanks

So the original area of the square is \(x^2\) and after the increase, it is \(0.6x*1.5x\), which equals \(0.9x^2\). Dividing the two areas, you see there is a 10% decrease. 

P.S. Sorry about the extra guest answers, my browser was glitching out. 

The area of the triangle is 1/2*6*6.5=19.5 centimeters squared.

The area of the circle is 4^2pi=16pi.

Thu,s the probability of a point landing inside the red shaded region is 19.5/16pi\(\approx\)0.388

Thanks CPhill and Cal 

Thanks, tertre  !!!!

So the original area of the square is \(x^2\) and after the increase, it is \(0.6x*1.5x\), which equals \(0.9x^2\). Dividing the two areas, you see there is a 10% decrease. 

This site will help:

So the original area of the square is \(x^2\) and after the increase, it is \(0.6x*1.5x\), which equals \(0.9x^2\). Dividing the two areas, you see there is a 10% decrease. 

Area of circle  = pi * 4^2  = 16pi   (1) 

Area of triangle  = (1/2)(base) (height)  =  (1/2)(6) ( 2.5 + 4)  =  3 * ( 6.5)  =  19.5     (2)

Prob that a point falls in the triangle =  (2) / (1)  =    19.5  / [ 16pi]  = .3879   =  .388

This is not the exact question, but it will help!

Since line A is parallel to line y=2x+4, line A must have the same slope which is two(2).

y=2x+b

Plug in (2,5) for x and y respectively

5=2(2)+b, 5=4+b, b=1

Thus, the y-intercept of line A is one(1).

Hint:

The total score will add up to: 81*5=405

Add up the first three: 231

Hence, the added value of the two test scores will be 174. Now, find the value of the two scores :D

Hope this helped!