50) Four circles of equal radii are centered at the four vertices of a square. These 4 circles touch a fifth circle of
equal radius placed inside this square. The ratio of the shaded area of the circles to the un-shaded area of the
circles is.
A) 5:2
B) 2:3
Ø 3:2
D) 2:5

The ratio of the shaded area of the circles to the un-shaded area of the  circles is 2:3. Option B is correct.

The formula for calculating the area of a circle is expressed as:

A = πr²

Let the area of the shaded part be and the area of the unshaded part be

The area of the shaded part is 2 circles while the area of the total unshaded part is 3 circles.

and

Taking the ratios of the areas

Hence the ratio of the shaded area of the circles to the un-shaded area of the  circles is 2:3

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B. 2:3

Step-by-step explanation:

It's easier to see when counting by quarters of a circle instead of a whole circle.

We start with the shaded. We know the fully shaded one has 4 quarters and we can count 4 more surrounding it. That's a total of 8. Since there are 4 quarters in each circle, we can divide it by 4 to get 2 full circles that are shaded.

Next is the white, they come in groups of 3 quarters since one of the quarters are shaded. There are 4 groups of these so we can do 3x4 to figure out the total number (or just count them out). We get a total of 12 white quarters. Divide by 4, we have 3 full white circles.

The ratio is 2:3

the image of (6, 9) under a dilation is (4, 6). the scale factor is

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12