Jenna's dogs used 9 99 chew toys last year. The animal shelter used 2 , 277 2,2772, comma, 277 chew toys. How many times as many chew toys did the animal shelter use? Choose 1 Choose 1 (Choice A) A The animal shelter used 20 , 493 20,49320, comma, 493 times as many chew toys, because 9 × 2 , 277 = 20 , 493 9×2,277=20,4939, times, 2, comma, 277, equals, 20, comma, 493. (Choice B) B The animal shelter used 2 , 053 2,0532, comma, 053 times as many chew toys, because 2 , 277 ÷ 9 = 2 , 053 2,277÷9=2,0532, comma, 277, divided by, 9, equals, 2, comma, 053. (Choice C) C The animal shelter used 253 253253 times as many chew toys, because 2 , 277 ÷ 9 = 253 2,277÷9=2532, comma, 277, divided by, 9, equals, 253.

253

Step-by-step explanation:

Given

--- Jenna's dog

-- Animal Shelter

See comment for proper format of the question

Required

How many times did A use than J

The options are not clear. So, the general solution is as follows.

To solve this, we simply divide the chew toys used in A by the chew toys used by J

So, we have:

This gives:

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

68719476736

step-by-step explanation:

8*8*8*8*8*8*8*8*8*8*8*8=68719476736