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3 years ago

5

The perimeter it a square field is 224. How long is

1 answer:

geniusboy [140]3 years ago

8 0

Since each side of a sqare is equal and there are four sides to the square
$224 = 4x$
X being the side length.
Then you would simiply divide by 4 to isolate x
$x = 56$
A side length is 56

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3. How many solutions does the system of equations have?

miskamm [114]

Answer:

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Step-by-step explanation:

5 0

3 years ago

Find a closed-form solution to the integral equation y(x) = 3 + Z x e dt ty(t) , x > 0. In other words, express y(x) as a fun

MrMuchimi

Answer:

$y{x} = \sqrt{7+2Inx}$

Step-by-step explanation:

$y(x)= 3 + \int\limits^x_e {dx}/ \, ty(t) , x>0}$

Let say; By y(x)= y(e)

we have;

$y(e)= 3 + \int\limits^e_e {dt}/ \, ty= 3+0$

Using Fundamental Theorem of Calculus and differentiating by Lebiniz Rule:

$y^{1} (x) = 0 + 1/ xy$

$y^{1} = 1/xy$

dy/dx = 1/xy

$\int\limits {y} \, dxy = \int\limits \, dx/x$

$y^{2}/2 Inx + C$

RECALL: y(e) = 3

$(3)^{2} / 2 = In (e) + C$

$\frac{9}{2} =In(e)+C$

$\frac{9}{2} - 1 = C$

$\frac{7}{2} = C$

$y^{2} / 2 = In x +C$

$y^{2} / 2 = In x +7/2$

MULTIPLYING BOTH SIDE BY 2 , TO ELIMINATE THE DENOMINATOR, WE HAVE;

$y^{2} = {7+2Inx}$

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8 0

3 years ago

Sam has a rectangular field whose length and breadth are in the ratio 8:7. He wants to fence the field with bamboo to make it an

USPshnik [31]

Answer:

Length = 200 m

Breadth = 175 m

Step-by-step explanation:

Perimeter of the rectangular field = Total cost ÷ rate per metre

= 12000 ÷ 16

= 750 m

Length : breadth = 8 : 7

Length = 8x

Breadth = 7x

Perimeter = 750

2*(length + breadth) = 750

2*(8x + 7x) = 750

2* 15x = 750

30x = 750

x = 750/30

x = 25

Length = 8x = 8*25 = 200 m

Breadth = 7*25 = 175 m

8 0

2 years ago

Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera

Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = $\frac{1}{2}(\text{Base})\times (\text{Height})$

Area of ΔADC = $\frac{1}{2}(\text{CD})(\text{AD})$

= $\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})$

= $\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)$

= $\frac{1}{2}(\sqrt{169-144})(12)$

= $\frac{1}{2}(5)(12)$

= 30 cm²

Area of equilateral triangle II = $\frac{\sqrt{3} }{4}(\text{Side})^2$

Area of equilateral triangle II = $\frac{\sqrt{3}}{4}(13)^2$

= $\frac{(1.73)(169)}{4}$

= 73.0925

≈ 73.09 cm²

Area of rectangle III = Length × width

= CF × CD

= 7 × 5

= 35 cm²

Area of trapezium EFGH = $\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})$

Since, GH = GJ + JK + KH

17 = $\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}$

12 = $\sqrt{81-x^2}+\sqrt{225-x^2}$

144 = (81 - x²) + (225 - x²) + 2 $\sqrt{(81-x^2)(225-x^2)}$

144 - 306 = -2x² + $2\sqrt{(81-x^2)(225-x^2)}$

-81 = -x² + $\sqrt{(81-x^2)(225-x^2)}$

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = $\frac{1}{2}(5+17)(9)$

= 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

= 237.09 cm²

7 0

3 years ago

(100 points!!!)

Sergio039 [100]

Answer:

this will probably help

Step-by-step explanation:

3 0

2 years ago