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Luden [163]
3 years ago
15

Three piles of 7 beans each are to be made from 10 red, 5 yellow, and 6 green beans. If all of the beans must be used and each s

tack must contain at least one bean of each color, then what is the maximum number of red beans that can be put in one of the stacks?
Mathematics
1 answer:
Ket [755]3 years ago
7 0

Answer:

The maximum number of red beans that can be put in one of the stacks is 5.

Step-by-step explanation:

First, Every pile is made with 7 beans, but they need to have at least one bean of each color, so we need to include in the pile minimum one yellow bean and one green bean. If we make a stack with this condition, the stack is going to have: 1 yellow bean, 1 green bean, and 5 red beans.

Additionally, it is necessary to know if there are enough red beans for the other piles. In this case, we have 10 red beans so if we have 5 red beans in one stack, the other two stacks have enough red beans to fulfill with the condition to have at least one red bean.

So, the maximum number of red beans that can be put in one of the stacks is 5.

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65,527

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1. SSS; Triangle ABC is congruent to Triangle DEF; sorry there's no triangle symbol on here.

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2 years ago
Tyger uses brick which are 7cm tall to build a tower.His final tower is 26 bricks high.How many cm tall is it?
iris [78.8K]
If the 26 bricks are each 7 cm, seems like multiplication or repeated addition would solve the problem. Multiplication would be faster. 26 x 7 = 182
182 cm tall.
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3 years ago
Read 2 more answers
3. In a single growing season at the Smith Family Orchard, the average yield per apple tree is 150 apples when the number of tre
g100num [7]

Answer:

<em><u>A.10000</u></em>

<em><u>B.25 more trees must be planted</u></em>

Step-by-step explanation:

⇒Given:

  • The intial average yield per acre y_{i} = 150
  • The initial number of trees per acre  t_{i} = 100
  • For each additional tree over 100, the average yield per tree decreases by 1 i.e , if the number trees become 101 , the avg yield becomes 149.
  • Total yield = (number of trees per acre)*(average yield per acre)

<em>A.</em>

⇒If the total trees per acre is doubled , which means :

total number of trees per acre t_{f} = 2*t_{i} = 200

the yield will decrease by : t_{f} - y_{i}

y_{f}= 150-100= 50

⇒total yield = 50*200=10000

<em>B.</em>

⇒to maximize the yield ,

let's take the number of trees per acre to be 100+y ;

and thus the average yield per acre = 150 - y;

total yield = (100+y)*(150-y)\\=15000+50y-y^{2} \\

this is a quadratic equation. this can be rewritten as ,

     ⇒   =15000+50y-y^{2}\\=15000+625 - (625 - 50y +y^{2})\\=15625 - (y-25)^{2}

In this equation , the total yield becomes maximum when y=25;

<u><em>⇒Thus the total number of trees per acre = 100+25 =125;</em></u>

                 

3 0
2 years ago
Find the product of all real values of r for which 1/2x=r-x/7
Dahasolnce [82]

Answer:

r = \±\sqrt{14

Product = -14

Step-by-step explanation:

Given

\frac{1}{2x} = \frac{r - x}{7}

Required

Find all product of real values that satisfy the equation

\frac{1}{2x} = \frac{r - x}{7}

Cross multiply:

2x(r - x) = 7 * 1

2xr - 2x^2 = 7

Subtract 7 from both sides

2xr - 2x^2 -7= 7 -7

2xr - 2x^2 -7= 0

Reorder

- 2x^2+ 2xr  -7= 0

Multiply through by -1

2x^2 - 2xr +7= 0

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

b^2 - 4ac = 0

For a quadratic equation

ax^2 + bx + c = 0

By comparison with 2x^2 - 2xr +7= 0

a = 2

b = -2r

c =7

Substitute these values in b^2 - 4ac = 0

(-2r)^2 - 4 * 2 * 7 = 0

4r^2 - 56 = 0

Add 56 to both sides

4r^2 - 56 + 56= 0 + 56

4r^2 = 56

Divide through by 4

r^2 = 14

Take square roots

\sqrt{r^2} = \±\sqrt{14

r = \±\sqrt{14

Hence, the possible values of r are:

\sqrt{14 or -\sqrt{14

and the product is:

Product = \sqrt{14} * -\sqrt{14}

Product = -14

8 0
3 years ago
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