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2 years ago

9

A spotlight can be adjusted to effectively light a circular area of up to 6 meters in diameter. To the nearest tenth, what is th

e maximum area that can be effectively lit by the spotlight?

1 answer:

levacccp [35]2 years ago

3 0

You would find the radius by halving the diameter. Then you use the equation
$\pi \: r { }^{2} =$
and you get about 28.2 meters.

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In ΔMNO, the measure of ∠O=90°, the measure of ∠M=74°, and OM = 5.4 feet. Find the length of NO to the nearest tenth of a foot.

torisob [31]

Answer:

140.6 feet

Step-by-step explanation:

gave up on deltamath

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2 years ago

Some identical cylinders each have a radius of

Anni [7]

Answer:

8 cylinders

Step-by-step explanation:

The surface area of a cylinder is given by the formula ...

A = 2πr(r +l) . . . . . for radius r and length l

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The cylinders Rufus has each have an area of ...

A = 2π(5 cm)(5 cm +4 cm) = 90π cm² ≈ 282.74 cm²

That means 2500 cm² of paint will cover ...

2500 cm²/(282.74 cm²/cylinder) ≈ 8.84 cylinders

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2 years ago

A company can manufacture x hundred items for a total cost of C=300+1200x-100x. Find x if the total cost is 3,000

Scilla [17]

C=300+1200x-100x. Find x if the total cost is 3,000
3,000 = 300 + 1200x -100x
3000 = 300 + 1100x
2700 = 1100x
2700÷1100=x
x=27/11
x = 2 5/11 Round to 3

5 0

2 years ago

pllllllls I need help right triangle and trigonometry special right triangles find the value of each variable

yanalaym [24]

$\huge {\mathfrak{Answer : }}$

we know,

$\boxed{\cos( \theta) = \dfrac{base}{hypotenuse}}$

so,

$\cos(30) = \dfrac{24}{x}$

$\dfrac{ \sqrt{3} }{2} = \dfrac{24}{x}$

$x = \dfrac{24 \times 2}{ \sqrt{3} }$

$x = 8 \sqrt{3} \times 2$

$x = 16 \sqrt{3}$

now,

$\boxed{ \tan( \theta) = \frac{perpendcular}{base} }$

$\tan(30) = \dfrac{y}{24}$

$\dfrac{1}{ \sqrt{3} } = \dfrac{y}{24}$

$y = \dfrac{24}{ \sqrt{3} }$

$y = 8 \sqrt{3}$

6 0

2 years ago

Researchers have concluded that a dry basin began to fill with water in 1880. the water level rose an average of 1.6 millimeters

docker41 [41]

Answer:

the domain is [0,129]
112 mm
1942

Step-by-step explanation:

1. The function is good for years 1880 to 2009, 0 to 129 years after 1880. Values of x can be anything in the domain [0, 129].

__

2. 1950 -1880 = 70. The year 1950 corresponds to x=70, so the function tells us the water level rose ...

f(70) = 1.6·70 = 112 . . . . . mm

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3. We want to find x when f(x) = 100. That will be the solution to ...

100 = 1.6x

100/1.6 = x = 62.5

Then 62.5 years after 1880 is year 1942.5. The water level was 100 mm higher than in 1880 in the year 1942.

3 0

2 years ago