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3 years ago

Perimeter of a ground is 6m. Find the cost of fencing it at the rate of rs 250per meter

Mathematics

1 answer:

kondaur [170]3 years ago

8 0

Answer:

to find the rate of fencing, you should find the perimeter of the rectangle.

2(l+b)

2(155+125)

2*180

360m

cost of fencing = 360*10

= Rs 3600

now to find the cost of fielding, find the area

(l*b)

155*125

19375 square mtr

cost of fielding = 19375*15

= Rs 290625

Step-by-step explanation:

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Solve the inequality, 6- 3b﹤10- 3b- 18.. please helpp

saw5 [17]

Answer:

There is no solution

Step-by-step explanation:

Let's solve your inequality step-by-step.

6−3b<10−3b−18

Step 1: Simplify both sides of the inequality.

−3b+6<−3b−8

Step 2: Add 3b to both sides.

−3b+6+3b<−3b−8+3b

6<−8

Step 3: Subtract 6 from both sides.

6−6<−8−6

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3 years ago

Write 632,016.08 in two other forms

Rina8888 [55]

600,000.00+30,000,00+2,000.00+10.00+6.00+.8

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3 years ago

2cos theta= -2root3 sin(-theta)

maks197457 [2]

Hello here is a solution :

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4 years ago

The area of the circular base of a cylinder is 36 Square units. The height of the cylinder is 2 units

coldgirl [10]

Answer:

We are asked to solve for the lateral area of the cylinder and the formula is shown below:

Lateral area = 2*pi*r*h

We are also given with the area of the base such as:

Area of the base = 36pi square units

Area of the base = pi*r²

Solving for radius:

36pi = pi*r²

r=6 inches

Solving for the lateral area when height is 2 units:

Lateral area = 2*pi*6*2

Lateral area = 24pi inches²

The answer is the letter "B" 24 square units

Step-by-step explanation:

8 0

3 years ago

PLEASE ANSWER WILL GIVE POINTS!!!!!!!!!!!

iris [78.8K]

Given:

The figure of a right triangle ABC.

To find:

The trigonometric ratios $\sin B, \tan A, \cos B$ .

Solution:

Using Pythagoras theorem,

$Hypotenuse^2=Perpendicular^2+Base^2$

$AB^2=AC^2+BC^2$

$(10)^2=(6)^2+BC^2$

$100-36=BC^2$

$64=BC^2$

Taking square root on both sides.

$\sqrt{64}=BC$

$8=BC$

So, measure of BC is 8 units.

Now,

$\sin \theta=\dfrac{Opposite}{Hypotenuse}$

$\sin B=\dfrac{AC}{AB}$

$\sin B=\dfrac{6}{10}$

$\sin B=\dfrac{3}{5}$

Similarly,

$\tan \theta=\dfrac{Opposite}{Adjacent}$

$\tan A=\dfrac{AC}{BC}$

$\tan A=\dfrac{6}{8}$

$\tan A=\dfrac{3}{4}$

And,

$\cos \theta=\dfrac{Adjacent}{Hypotenuse}$

$\cos B=\dfrac{BC}{AB}$

$\cos B=\dfrac{8}{10}$

$\cos B=\dfrac{4}{5}$

Therefore, the required trigonometric ration are $\sin B=\dfrac{3}{5}, \tan A=\dfrac{3}{4}, \cos B=\dfrac{4}{5}$ .

6 0

3 years ago

Perimeter of a ground is 6m. Find the cost of fencing it at the rate of rs 250per meter​

Perimeter of a ground is 6m. Find the cost of fencing it at the rate of rs 250per meter