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DaniilM [7]
2 years ago
14

Approximate 59 pi to the nearest hundredth. SHOW WORKINGS PLZ!!!!

Mathematics
1 answer:
Neko [114]2 years ago
8 0
Ok here we go!!!...
When looking at pi make sure that you don't just plug in 3.14 because you will not get the exact answer. To get the exact answer you can use the pi symbol on your calculator.
This is what you will plug into your calculator....
59×π
The calculator will give you the result of 185.3539666
But, we are going to round to the nearest hundredth therefore we will only put the two numbers after the decimal.
Our final answer will be 185.35 because rounding from 185.353 we will round down and get 185.35 as our answer.

Hope I helped!!
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A farmer has 120 feet of
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you could fence each side with 40 feet if fence if the side with the barn does not need fencing

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3 years ago
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Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0​
IRISSAK [1]

<u>Prove that:</u>

\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0

<u>Proof: </u>

We know that, by Law of Cosines,

  • \sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}
  • \sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}
  • \sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}

<u>Taking</u><u> </u><u>LHS</u>

\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)

\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)

\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)

\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)

\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}

<em>On combining the fractions,</em>

\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}

\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}

<em>Regrouping the terms,</em>

\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}

\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}

\longmapsto\dfrac{0}{2abc}

\longmapsto\bf 0=RHS

LHS = RHS proved.

7 0
2 years ago
32 (23 + 4),
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Answer:

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at the county fair , 12% of the pepole who attended received free water bottels . if 1,500 people receive dfree water bottles ,
sweet-ann [11.9K]

12500 people went to fair

<h3><u>Solution:</u></h3>

Given that at the county fair , 12% of the people who attended received free water bottles

1,500 people received free water bottles

To find: number of peoples went to fair

Let "a" be the number of peoples went to fair

From given information,

12% of the people who attended received free water bottles and 1,500 people received free water bottles

Which means 12% of total people is equal to 1500

12% of total people = 1500

12 % of "a" = 1500

\frac{12}{100} \times a = 1500\\\\a = 1500 \times \frac{100}{12}\\\\a = \frac{150000}{12} = 12500

Thus 12500 people went to fair

3 0
2 years ago
Solve the inequality for x. Show each step of the solution. <br><br> 12x&gt;3(2x+4)-15
Leto [7]

Answer:

x<-2

Step-by-step explanation:

12x>6x+12-15 (first distribute)

12x>6x-3 (simplify)

6x>-3 (put like terms on one side by subtraction)

x<-2 (divide and switch inequality sign since you divided by a negative)

5 0
3 years ago
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