A Triangular Park ABC has sides 120 m, 80m and 50m. a gardener has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of of Rs. 20 per metre leaving a space 3m wide for a gate on one side?​

Answers

Answer 1

The perimeter = sum of all sides

= 120 + 80 + 50

= 250

So 250 - 3

247

Left space for gate

Now cost of fencing = Rs 20/per meter

= 247 × 20

= Rs 4,940

Now the area of the triangular park can be found using heron's formula

S = (a+b+c)/2

S = (120+80+50)/2

S = 250/2

S = 125

Now

Herons formula = √s(s-a)(s-b)(s-c)

√125(125-120)(125-80)(120-50)

√125(5)(45)(70)

√5×5×5×5×5×3×3×5×14

After Making pairs

5×5×5×3√14

375√14

Therefore 375√14m is the area of the triangular park

Must click thanks and mark brainliest

Answer 2

$\sf\underline\bold{Answer:}$

$\sf\small\underline{\underline{Area\: planted\: by\: the\: gardener : 1452.36m^2}}$

$\sf\small\underline{\underline{The\:cost\:of\:fencing\:the\:park:Rs.4940}}$

$\space$

$\sf\underline\bold{Step-by-Step:}$

$\space$

$\sf\bold{Given(In \:the\:Q):}$

Sides of the triangular park are 120m,80m and 50m.

$\space$

$\sf\bold{To \: find:}$

How much area of the park does she need to plant?The cost of fencing the park ?

$\space$

$\sf\small{☆Area\:to\:be\:planted=Area \: of \: ∆ABC}$

$\space$

$\sf\underline\bold{Calculating\:area\:of\:∆ABC:}$

$\space$

Use heron's formula to find the area of the triangle.

$\space$

$\mapsto$ $\sf\small{Heron's\:formula=}$

[tex]\sf\sqrt{s(s-a)(s-b)(s-c)}[/tex]

$\sf{Where\:s=semi\:perimeter}$$\sf{a,b,c\: = side\:of\:the\:∆}$$\sf\small{Here\:a=120,b=80 \:and\: c=50}$

$\space$

$\sf\bold{Now,find\:semi\:parameter:-}$

$\sf\small{Perimeter\:of\:the\:∆=120+80+50=250}$

$\sf\small{Semi-Perimeter:}$ $\sf\dfrac{250}{2}$ $\sf\small{=125m}$

$\space$

$\sf\small{Substitute \: the\:values\:in\:heron's\:formula:}$

$\sf{Area\:of\:the\:∆:-}$

$\mapsto$ [tex]\sf\sqrt{125(125-120)(125-80)(125-50)}[/tex]

$\space$

$\mapsto$ $\sf\sqrt{125\times(5)\times(45)\times(75)}$

$\space$

$\mapsto$ $\sf\small\sqrt{2109375}$ $\sf\small{=375}$ $\sf\small\sqrt{15}$

$\space$

$\longmapsto$ $\sf\underline\bold\purple{1452.56m^2}$

______________________________

$\sf\underline\bold{Now,find\:the\:cost\:of\:fencing:}$

$\sf{Cost\:of\:fencing-}$

$\sf{Rate = Rs.20 per \:meter}$ $\sf{Left\:space=3m}$

$\space$

$\sf\underline{Hence,the\:gardener\:has\:to\:fence:}$

$\sf{= 250-3=247m.}$

$\space$

So,total cost of fencing at the rate of Rs.20 per m:-

$\sf\underline\bold\purple{=247\times20=4940}$

___________________________________


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Phone calls to texts is 3 to 5, so for every 3 phone calls she makes 5 texts:

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Answers

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Answers

Lets solve

[tex]\\ \sf\longmapsto 7w<-56[/tex]

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Answer:

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Answers

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Answers

Answer:

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Step-by-step explanation:

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Answers

The answer is 81 < < 105

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Answers

Answer:

$17,000 at 14% and $9,000 at 2%

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Let x = amount invested at 14%.

Let y = amount invested at 2%.

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-------------------------------------

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Combine like terms.

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Answer:

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Answer:

z^2-3z-18

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-19>t

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Answers

y=2

PREMISES

y=the expression 19+7/2×6+1 rewritten by including two sets of brackets or parentheses to get a result of 2

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To clearly express the mathematical phrase 19+7/2×6+1 to get a result of 2 requires specific parenthetical treatment.

y=19+7/2×6+1 becomes

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TO BE ANSERED ASAP
If n = 4, then 9ˆ8 ÷ 9 n is equal to__________________.

Answers

Answer:

1,195,742.25 (final answer since instructions did not include rounding up or anything)

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n = 4

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9^8 ÷ 9n

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9^8 ÷ 9n

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43046721 ÷ 9(4)

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= 1,195,742.25

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[tex]( - 12) \times ( - 4) + ( - 8) + ( + 3)[/tex]
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Answers

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The business college computing center wants to determine the proportion of business students who have personal computers (PC's) at home. If the proportion exceeds 30%, then the lab will scale back a proposed enlargement of its facilities. Suppose 300 business students were randomly sampled and 65 have PC's at home. What assumptions are necessary for this test to be satisfied

Answers

Solution :

Given data :

x = 65, n = 300

[tex]$\hat p = \frac{x}{n}[/tex]

  [tex]$=\frac{65}{300}$[/tex]

  = 0.2167

The hypothesis are :

[tex]$H_0: p \leq 0.3$[/tex]

[tex]$H_0: p> 0.3$[/tex]

The [tex]\text{level of significance}[/tex], α = 0.05

The test is right tailed.

The standard deviation is :

[tex]$\sigma = \sqrt{\frac{0.3(1-0.3)}{300}}$[/tex]

σ = 0.0265

The test statistics is :

[tex]$z=\frac{\hat p - p}{\sigma}$[/tex]

[tex]$z=\frac{0.2167 - 0.3}{0.0265}$[/tex]

  = -3.14

The critical value is 1.645

The rejection region is : If z > 1.645, then we reject [tex]H_0[/tex]

Decision :

Since the test statistics does not lie in the rejection, so we fail to . [tex]\text{reject the null hypothesis}[/tex].

P-value : [tex]$P(z > - 3.14)$[/tex]   = 0.9992

Therefore, the p-value is not less than the level of significance so we fail to [tex]\text{reject the null hypothesis}[/tex].

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