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gogolik [260]
3 years ago
6

A lawn sprinkler spins in a circle. The sprinkler covers a radius of 4 feet. Which choice is the closest to the area of lawn tha

t the sprinkler can cover? 13, 25, 50, or 200
I think its 50, am I right?
Mathematics
2 answers:
KiRa [710]3 years ago
8 0
Yes u r right 50 is the correct answer after u do all the required math. U can also figure it out by seeing that it only covers a 4 feet radius then u can pretty much just tell the answer of the asked question and hope u pass have a great day.
konstantin123 [22]3 years ago
5 0

Answer: Yes, the closest to the area of lawn that the sprinkler can cover= 50\ ft^2


Step-by-step explanation:

Given: The radius of the area covered by sprinkler= 4 feet

We know that the area of a circle with radius r is given by,

Area=\pi\ r^2\\\Rightarrow\ Area=3.14\times4^2\\\Rightarrow\ Area=3.14\times16=50.24\ ft^2\\\

Hence, the area covered by sprinkler= 50.24\ ft^2

Now, the closest to the area of lawn that the sprinkler can cover= 50\ ft^2

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solve the eqution 9.95+0.30s=c c is the amount of dollars a website charges for downloading songs s stands ffor number of songs
lutik1710 [3]

9.95 + 0.30s = C

9.95

9.95 + 10.5  = C

C = 20.45

hope this helps!

~LEAN~

5 0
3 years ago
What is the circumference of a 12-inch diameter pizza? Leave the answer in terms of ?.
adoni [48]

Answer:

C = 37.7 inches

Step-by-step explanation:

You are looking for the circumference.

Formula for the circumference-> C=2πr

variables:

c = circumference

r = radius

You have the diameter, which is 12 inches. Radius is half of the diameter. which means r is half of 12 (aka 6).

put the radius where the variable is in the formula.

C=2π(6)

you now have enough information to solve for circumference.

C = 2 * π * 6

C = 37.7

4 0
3 years ago
One survey estimates that,on average, the retail value of a mid sized car decreases by8% annually.if the retail value of a car i
Scrat [10]
Percentage by which the average value of mid sized car decreases each year = 8%
Retail value of a car today = v dollars
Amount of decrease in the value of the car after 1 year = (8/100) * v
                                                                                         = 2v/25 dollars
Then
The equation that represents the value of the car after 1 year = v - (2v/25) dollars
                                                                                                   = (25v - 2v)/25 dollars
                                                                                                   = 23v/25 dollars
So following this expression the value of the mid sized car can be easily determined after 1 year. I hope this is the answer you were looking for and the procedure is also clear to you.
5 0
3 years ago
Read 2 more answers
Giúp tớ giải bài toán này với
vampirchik [111]

Answer:

english pls

Step-by-step explanation:

6 0
2 years ago
Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane
liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

Answer:

Rate = 0.935042^\circ /cm

Step-by-step explanation:

Given

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

T(x,y) =x\sin2y

r = 1m

v = 2m/s

Express the given point P as a unit tangent vector:

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

u = \frac{\sqrt 3}{2}i - \frac{1}{2}j

Next, find the gradient of P and T using: \triangle T = \nabla T * u

Where

\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})}  = (sin \sqrt 3)i + (cos \sqrt 3)j

So: the gradient becomes:

\triangle T = \nabla T * u

\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] *  [\frac{\sqrt 3}{2}i - \frac{1}{2}j]

By vector multiplication, we have:

\triangle T = (sin \sqrt 3)*  \frac{\sqrt 3}{2} - (cos \sqrt 3)  \frac{1}{2}

\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)

\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5

\triangle T = 0.935042

Hence, the rate is:

Rate = \triangle T = 0.935042^\circ /cm

3 0
2 years ago
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