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Masja [62]
3 years ago
15

I need to understand question 14 and 15 and the last one cut off for tens in one

Mathematics
1 answer:
sp2606 [1]3 years ago
6 0

                                                 Question # 14

Given the numbers

10     11     12     13     14     15     16     17     18     19     20

Let 'x' be the number

The condition breakdown:

I am less than 20.

  • So the number 'x' must be less than 20 i.e. x < 20

I am more than 13.

  • So the number 'x' must be greater than 13 i.e. x > 13

I am less than 17.

  • So the number 'x' must be less than 17 i.e. x < 17

Finally:

I am 4 more than 12

i.e. 12+4 = 16

Thus, the number is x = 16

                                               Question # 15

Part a)

Given the numbers

10     11     12     13     14     15     16     17     18     19     20

Let 'x' be the number

The condition breakdown:

I am more than 10.

  • x > 10

I am less than 20.

  • x < 20

I am more than 12.

  • x > 12

I am less than 15.

  • x < 15

As the numbers left after all the conditions are fulfilled are 13 and 14.

  • But the last condition is, of the numbers that left, the number is greater than all the remaining numbers.

So, from the remaining number 13 and 14;

14 > 13

Thus, the number x = 14

Part b)

Drawing the number 14 in the place value:

          Chart

Tens                Ones

1                         4

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\green{\large\underline{\sf{Solution-}}}

<u>Given expression is </u>

\rm :\longmapsto\:\dfrac{5 \times  {25}^{n + 1}  - 25 \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {25}^{n + 1} }

can be rewritten as

\rm \:  =  \: \dfrac{5 \times  { {(5}^{2} )}^{n + 1}  -  {5}^{2}  \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {( {5}^{2} )}^{n + 1} }

We know,

\purple{\rm :\longmapsto\:\boxed{\tt{  {( {x}^{m} )}^{n}  \: = \:   {x}^{mn}}}} \\

And

\purple{\rm :\longmapsto\:\boxed{\tt{ \:  \:   {x}^{m} \times  {x}^{n} =  {x}^{m + n} \: }}} \\

So, using this identity, we

\rm \:  =  \: \dfrac{5 \times  {5}^{2n + 2}  - {5}^{2n + 2} }{{5}^{2n + 3 + 1}  -  {5}^{2n + 2} }

\rm \:  =  \: \dfrac{{5}^{2n + 2 + 1}  - {5}^{2n + 2} }{{5}^{2n + 4}  -  {5}^{2n + 2} }

can be further rewritten as

\rm \:  =  \: \dfrac{{5}^{2n + 2 + 1}  - {5}^{2n + 2} }{{5}^{2n + 2 + 2}  -  {5}^{2n + 2} }

\rm \:  =  \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2}  - 1)}

\rm \:  =  \: \dfrac{4}{25 - 1}

\rm \:  =  \: \dfrac{4}{24}

\rm \:  =  \: \dfrac{1}{6}

<u>Hence, </u>

\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times  {25}^{n + 1}  - 25 \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {25}^{n + 1} }  =  \frac{1}{6} }}

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2 years ago
A fence must be built to enclose a rectangular area of 45 comma 000 ftsquared. Fencing material costs $ 3 per foot for the two s
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Answer:

150 feet by 300 feet.

Step-by-step explanation:

The fence is to enclose a rectangular area of 45,000 ft squared.

If the dimensions of the rectangle are x and y

Area of a rectangle = xy

  • xy=45000
  • x=\frac{45000}{y}

Perimeter of the Rectangle =2x+2y

Fencing material costs $ 3 per foot for the two sides facing north and south and ​$6 per foot for the other two sides.

  • Cost of Fencing, C=$(6*2x+3*2y)=$(12x+6y)

Substitute x=\frac{45000}{y} into the Cost to get C(y)

C=12x+6y

C(y)=12(\frac{45000}{y})+6y\\C(y)=\frac{540000+6y^2}{y}

The value at which the cost is least expensive is at the minimum point of C(y), when the derivative is zero.

C^{'}(y)=\dfrac{6y^2-540000}{y^2}

\dfrac{6y^2-540000}{y^2}=0\\6y^2-540000=0\\6y^2=540000\\y^2=\frac{540000}{6} =90000\\y=\sqrt{90000}=300

Recall,

x=\frac{45000}{y}=\frac{45000}{300}=150

Since x=150, y=300

The dimensions that will be least expensive to build is 150 feet by 300 feet.

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