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olga_2 [115]
3 years ago
12

Christine is going to use 20m of fencing to fence off a rectangular field to grow vegetables. She is considering three fields wi

th lengths of 3m, 5m, and 6m.
Answer the questions below to find which of these fields would hold the most vegetables.
Mathematics
1 answer:
Pie3 years ago
6 0

Answer:

6m field will hold most vegetables

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A movie theater sell snacks to 30% of it's customer each day. On Saturday, 186 customers bought snacks. How many customers went
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620 or D

Step-by-step explanation:

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A combination lock has 6 different numbers. You can use any number 0 - 9 in the combination, but each number can only be used ON
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There are 151,200 possible combinations. If you'd like an explanation of my answer, please let me know.
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Assume that 0 a) 25/24 b) -25/24 c) 24/25 d) -24/25
ser-zykov [4K]
A) 25/24 because i don’t know i’m just doing this to get points
8 0
3 years ago
What is f[g(3)] for the following function? <br> f(x) = 4x2 − 3<br><br> g(x) = 5x − 2
lukranit [14]

Answer:

673

Explanation:

  • f(x) = 4(x)² − 3
  • g(x) = 5(x) − 2

solve:

  • f[g(3)]
  • f(5(3) − 2)
  • f(13)
  • 4(13)² − 3
  • 673
8 0
2 years ago
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Hey guys<br>im new here<br>please solve this for me with steps!<br>ill mark as the best answer​
Vinil7 [7]

Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

5 0
3 years ago
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