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3 years ago

9

Write the equation of the trigonometric graph.

1 answer:

prohojiy [21]3 years ago

5 0

It need to be bigger

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Prove the identity (n-5)^2 - (6n-35)=(n-10)(n-6)

elena-14-01-66 [18.8K]

Answer:

(n-5)^2 - (6n-35)=(n-10)(n-6)

-----------

n²-10n+25-6n+35 =
n²-16+60 = n²- 10n - 6n + 60 =
n(n-10) - 6(n-10) =
(n-10)(n-6)

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3 years ago

A school is designed so that 5 hallways meet at the library in the shape of a regular pentagon. Find the measure of one of the e

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4 years ago

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The diameter of a circle is 22 miles. What is the circle's area?

Elodia [21]

Answer:
379.94

Explanation :
check out my attached work
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Amy is using a drawing program to complete a construction with which she is almost finished. Which construction is she completin

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4 years ago

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A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. The diameter at the base

timurjin [86]

Answer:

Step-by-step explanation:

Equation for a hyperboloid of one sheet, with center at the origen and axis along z-axis is:

(x/a)² + (y/b)² - (z/c)² = 1 (1)

We have to find a , b, and c

We can express equation (1)

(x/a)² + (y/b)² = (z/c)² + 1 (2)

Now if we cut the hyperboloid with planes parallel to xy plane we get for z = k ( K = 1 , 2 , 3 and so on ) circles of different radius

(x/a)² + (y/b)² = (k/c)² + 1

at z = k = 0 at the base of the hyperboloid d = 300 or r = 150 m

we have

(x/a)² + (y/b)² = 1

x² + y² = a² a² = (150)² a = b = 150

and x² + y² = (150)²

Now the other condition is at 200 m above the base d = 500 m r = 250 m minimum diameter then in equation (2) we have:

(x/a)² + (y/b)² = (z/c)² + 1

(1/a)² [ x² + y² ] = (z/c)² + 1

but x² + y² = r² and in this case r = 250 m then

(250)²/(150)² = (z/c)² + 1 ⇒ (62500/ 22500) = (200/c)² + 1

2,78 = 40000/c² + 1

2.78c² = 40000 + c²

1.78c² = 40000

c² = 40000/1.78

c² = 22471.91

c = 149,91

Then we finally have the equation:

x²/(150)² + y² /(150)² - z²/149,91 = 1

7 0

4 years ago