The Square sand pit in park lot has an area of 16 ft. There is a rope along one side of the pit. How long is the rope?

Find the values of x and y to make the equation true.<br> -2x + 61 = (-24y)i - 14

gregori [183]

X= 75/2
Y=0
That is the answer to your question I used Photomath to help me

4 0

2 years ago

Help me on 1,2,3,5,and 6 please

arlik [135]

Since most of these are mixed number you have to convert them into improper fractions.
For an example 2 3/4 x 1/2
First you would have the convert 2 3/4
You would add 2 to 3 then multiply 2 to 4
So it would be (2x4) + 3
You keep the denominator of the originally fraction so it would be 11/4
Then to finish you would multiply straight across so
11x1= 11
4x2=8
11/8
1 3/8
Hope this helps!!

8 0

3 years ago

I need to solve for X<br> X=

Gemiola [76]

$\dfrac{2\frac{1}{4}}{\frac{3}{8}}=\dfrac{\frac{15}{16}}{x}\ \ \ \ |\text{cross multiply}\\\\2\dfrac{1}{4}x=\dfrac{3}{8}\cdot\dfrac{15}{16}\\\\\dfrac{2\cdot4+1}{4}x=\dfrac{45}{128}\\\\\dfrac{9}{4}x=\dfrac{45}{128}\ \ \ \ \ |\text{multiply both sides by 4}\\\\9x=\dfrac{45}{32}\ \ \ \ \ |\text{divide both sides by 9}\\\\\boxed{x=\dfrac{5}{32}}$

6 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

The isosceles triangle shown has a base of 4 y + 2 and a perimeter of 6 y + 12 . Which expression represents the length of 1 o

liberstina [14]

Answer:

The expression that represents the length of 1 of the triangle's legs is y + 5

Step-by-step explanation:

An isosceles triangle has two sides equal which are the triangle legs. Let b represent the base of the triangle and l represent one of the triangle's legs. Then, the perimeter, P is given by

P = l + l + b

i.e P = 2l + b

From the question, P = 6y + 12 and b = 4y +2

∴ 6y + 12 = 2l + 4y + 2

6y - 4y + 12 - 2 = 2l

2y + 10 = 2l

∴ 2l = 2y + 10

Then,

l = (2y+10)/2

l = y + 5

Hence, the expression that represents the length of 1 of the triangle's legs is y + 5

4 0

2 years ago