Arthur wanted to estimate the value of $52.3 \div 0.11$, so he decided to round each of the numbers 52.3 and 0.11 to the nearest whole number before dividing. arthur's plan failed badly. what would have been his estimate if he had rounded each of the numbers to its biggest decimal place (that is, 52.3 to the nearest ten and 0.11 to the nearest tenth) before dividing?

500.

Step-by-step explanation:

0.11 to the nearest whole number  is 0 so that failed because you cant divide by 0.

52.3 to the nearest ten is 50 and 0.11 to the nearest tenth  is 0.1 so the estimate would be 50 / 0.1

= 500 / 1

= 500.

Solution:

Rounding 52.3 to the nearest ten gives 50, while rounding 0.11 to the nearest tenth gives 0.1. Dividing 50 by 0.1 is the same as multiplying 50 by the reciprocal of 0.1, which is 10. So 52.3 ÷ 0.11 can be estimated as 50 × 10 = 500.

So the answer is 500!!!

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