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garik1379 [7]
3 years ago
15

74.21 in written form

Mathematics
1 answer:
DENIUS [597]3 years ago
5 0
Seventy four and twenty-one hundredths. 
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\bf{1) \  x^4-61x^2+900=0 }

Applying the factorization method, for example:

      \bf{ 0=(x^2)^2-61(x^2)+900=(x^2-36)(x^2-25)}

               \bf{\\ &=(x+6)(x-6)(x+5)(x-5);    }

              \bf{ x_1=6 \quad x_2=-6 \quad \ \ \ x_3=5 \quad x_4=5  }

\bf{2) \  x^4-25x^2+144=0  }

Applying the factorization method:

          \bf{0&=(x^2)^2-25(x^2)+144=(x^2-16)(x^2-9) }

             \bf{ =(x-4)(x+4)(x-3)(x+3); }

            \bf{ x_1=4 \quad x_2=-4 \quad x_3=3 \quad x_4=-3  }

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What is the y-intercept of the line shown?
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In the diagram below, what kind of line is BF relative to ABC?
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Step-by-step explanation:

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The yield of orange trees is reduced if they are planted too close together. If there are 30 trees per acre, each tree produces
Tanya [424]

Answer:

5 more than 30 trees should be  planted, for a total of 40 trees per acre.

Step-by-step explanation:

Let x be the number of trees beyond 30 that are planted on the acre

The number of oranges produced = Oranges(x) = (number of trees) (yield per tree)

We are given that For each additional tree in the acre, the yield is reduced by 7 oranges per tree

So, number of oranges produced =(30 + x)(400 -10x)

                                                        = 12000 -300x+400x-10x^2

                                                        = 12000 + 100x-10x^2

The derivative Oranges'(x) = 100-20x

Substitute first derivative equals to 0

100-20x =0

x=5

Using the second derivative test,

Oranges"(x) = -20

20 is negative,

So, this is the case of maximum.

Thus, 5 more than 30 trees should be  planted, for a total of 40 trees per acre.

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